( 20 ) ∫ s i n 2 x c o s 3 x d x = ∫ c o s − 3 x s i n 2 x d x ，根据积分表（十一）公式 99 和 98 ，得 ∫ c o s − 3 x s i n 2 x d x = c o s − 2 x s i n x − ∫ c o s − 3 x d x = c o s − 2 x s i n x − 1 2 c o s − 2 x s i n x − 1 2 ∫ 1 c o s x d x = 1 2 c o s − 2 x s i n x − 1 2 l n ∣ s e c x + t a n x ∣ + C = 1 2 s e c x t a n x − 1 2 l n ∣ s e c x + t a n x ∣ + C ( 21 ) 设 u = a r c t a n x ，则 x = t a n 2 u ， d x = 2 t a n u s e c 2 u d u ，得 ∫ a r c t a n x d x = 2 ∫ u t a n u s e c 2 u d u = 2 ∫ u s e c u d ( s e c u ) ，设 s = s e c u ， d s = d ( s e c u ) ，则 u = a r c c o s 1 s ， 得 2 ∫ u s e c u d ( s e c u ) = 2 ∫ a r c c o s 1 s ⋅ s d s ，令 u = a r c c o s 1 s ， d v = s d s ，则 d u = − s s 2 − 1 d s ， v = 1 2 s 2 ，得 2 ∫ a r c c o s 1 s ⋅ s d s = s 2 a r c c o s 1 s + ∫ s 3 s 2 − 1 d s ，根据积分表（八）公式 66 得， s 2 a r c c o s 1 s + ∫ s 3 s 2 − 1 d s = s 2 a r c c o s 1 s − s 2 − 1 + C = ( 1 + x ) a r c t a n x − x + C ( 22 ) ∫ 1 + c o s x s i n x d x = ∫ 2 ∣ c o s x 2 ∣ 2 s i n x 2 c o s x 2 d x = ± 2 ∫ c s c x 2 d ( x 2 ) = ± 2 l n ∣ c s c x 2 − c o t x 2 ∣ + C 当 c o s x 2 > 0 时， l n ∣ c s c x 2 − c o t x 2 ∣ = l n ( ∣ c s c x 2 ∣ − ∣ c o t x 2 ∣ ) 当 c o s x 2 < 0 时， l n ∣ c s c x 2 − c o t x 2 ∣ = l n ( ∣ c s c x 2 ∣ + ∣ c o t x 2 ∣ ) = − l n ( ∣ c s c x 2 ∣ − ∣ c o t x 2 ∣ ) 因此， ∫ 1 + c o s x s i n x d x = 2 l n ( ∣ c s c x 2 ∣ − ∣ c o t x 2 ∣ ) + C ( 23 ) ∫ x 3 ( 1 + x 8 ) 2 d x = ∫ x 4 x ( 1 + ( x 4 ) 2 ) 2 d x ，设 u = x 4 ， x = x 4 ， d x = − 1 4 u 3 4 d u ，得 ∫ x 3 ( 1 + x 8 ) 2 d x = ∫ u u 4 ( 1 + u 2 ) 2 ⋅ 1 4 u 3 4 d u = 1 4 ∫ 1 ( 1 + u 2 ) 2 d u ，根据积分表（四）公式 28 ，得 1 4 ( u 2 ( u 2 + 1 ) + 1 2 ∫ 1 1 + u 2 d u ) = u 8 ( u 2 + 1 ) + 1 8 a r c t a n u + C = x 4 8 ( x 8 + 1 ) + 1 8 a r c t a n x 4 + C ( 24 ) 设 u = x 4 ， x = u 4 ，则 d x = 1 4 u 3 4 d u ，得 ∫ x 11 x 8 + 3 x 4 + 2 d x = ∫ u 11 4 u 2 + 3 u + 2 ⋅ 1 4 u 3 4 d u = 1 4 ∫ u 2 u 2 + 3 u + 2 d u = 1 4 ∫ ( 1 + 1 u + 1 − 4 u + 2 ) d u = 1 4 u + 1 4 l n ∣ 1 + u ∣ − l n ∣ 2 + u ∣ + C = 1 4 x 4 + l n 1 + x 4 4 2 + x 4 + C ( 25 ) ∫ d x 16 − x 4 = 1 8 ∫ ( 1 4 + x 2 + 1 4 − x 2 ) d x = 1 8 ∫ 1 4 + x 2 d x + 1 32 ∫ ( 1 2 + x + 1 2 − x ) d x = 1 16 a r c t a n x 2 + 1 32 l n ∣ 2 + x 2 − x ∣ + C ( 26 ) ∫ s i n x 1 + s i n x d x = ∫ 2 t a n x 2 1 + t a n 2 x 2 ( t a n x 2 + 1 ) 2 1 + t a n 2 x 2 d x = ∫ 2 t a n x 2 ( t a n x 2 + 1 ) 2 d x ，设 u = t a n x 2 ，则 x = 2 a r c t a n u ， d x = 2 1 + u 2 d u ，得 ∫ 2 t a n x 2 ( t a n x 2 + 1 ) 2 d x = 4 ∫ u ( u + 1 ) 2 ⋅ 1 1 + u 2 d u = 2 ∫ ( 1 1 + u 2 − 1 ( 1 + u ) 2 ) d u = 2 ∫ 1 1 + u 2 d u − 2 ∫ 1 ( u + 1 ) 2 d u = 2 a r c t a n u + 2 1 + u + C = x + 2 1 + t a n x 2 + C ( 27 ) ∫ x + s i n x 1 + c o s x d x = ∫ x + 2 s i n x 2 c o s x 2 2 c o s 2 x 2 d x = 2 ∫ x 2 s e c 2 x 2 d ( x 2 ) + ∫ t a n x 2 d x ， 设 u = x 2 ， d v = s e c 2 x 2 d ( x 2 ) ，则 d u = 1 2 d x ， v = t a n x 2 ，得 2 ∫ x 2 s e c 2 x 2 d ( x 2 ) + ∫ t a n x 2 d x = x t a n x 2 − ∫ t a n x 2 d x + ∫ t a n x 2 d x + C = x t a n x 2 + C ( 28 ) ∫ e s i n x x c o s 3 x − s i n x c o s 2 x d x = ∫ x e s i n x c o s x d x = ∫ e s i n x t a n x s e c x d x = ∫ x d ( e s i n x ) − ∫ e s i n x d ( s e c x ) 设 u = x ， d v = d ( e s i n x ) ，则 d u = d x ， v = e s i n x ，得 ∫ x d ( e s i n x ) = x e s i n x − ∫ e s i n x d x ， 设 u = e s i n x ， d v = d ( s e c x ) ，则 d u = e s i n x c o s x d x ， v = s e c x ，得 ∫ e s i n x d ( s e c x ) = s e c x e s i n x − ∫ e s i n x d x ，因此 ∫ x d ( e s i n x ) − ∫ e s i n x d ( s e c x ) = x e s i n x − ∫ e s i n x d x − s e c x e s i n x + ∫ e s i n x d x = ( x − s e c x ) e s i n x + C ( 29 ) 设 u = x 6 ，则 x = u 6 ， d x = 6 u 5 d u ，得 ∫ x 3 x ( x + x 3 ) d x = ∫ u 2 u 6 ( u 3 + u 2 ) ⋅ 6 u 5 d u = 6 ∫ 1 u ( u + 1 ) d u = 6 ∫ ( 1 u − 1 u + 1 ) d u = 6 l n ∣ u u + 1 ∣ + C = 6 l n x 6 x 6 + 1 + C = l n x ( x 6 + 1 ) 6 + C ( 30 ) 设 u = e x ，则 x = l n u ， d x = 1 u d u ，得 ∫ d x ( 1 + e x ) 2 = ∫ 1 u ( 1 + u ) 2 d u = ∫ ( 1 u − 1 1 + u − 1 ( 1 + u ) 2 ) d u = l n u − l n ( 1 + u ) + 1 1 + u + C = x − l n ( 1 + e x ) + 1 1 + e x + C ( 31 ) ∫ e 3 x + e x e 4 x − e 2 x + 1 d x = ∫ e x + e − x e 2 x − 1 + e − 2 x d x = ∫ 1 ( e x − e − x ) 2 + 1 d ( e x − e − x ) = a r c t a n ( e x − e − x ) + C ( 32 ) 设 u = e x ，则 x = l n u ， d x = 1 u d u ，得 ∫ x e x ( e x + 1 ) 2 d x = ∫ l n u ( 1 + u ) 2 d u ，令 s = l n u ， d v = 1 ( 1 + u ) 2 d u ，则 d s = 1 u d u ， v = − 1 1 + u ，得 ∫ l n u ( 1 + u ) 2 d u = − l n u 1 + u + ∫ 1 u ( 1 + u ) d u = − l n u 1 + u + ∫ 1 u d u − ∫ 1 1 + u d u = − l n u 1 + u + l n u − l n ( 1 + u ) + C = − x 1 + e x + x − l n ( 1 + e x ) + C ( 33 ) 设 u = l n 2 ( x + 1 + x 2 ) ， d v = d x ，则 u = 2 l n ( x + 1 + x 2 ) 1 + x 2 d x ， v = x ，得 ∫ l n 2 ( x + 1 + x 2 ) d x = x l n 2 ( x + 1 + x 2 ) − ∫ 2 x l n ( x + 1 + x 2 ) 1 + x 2 d x = x l n 2 ( x + 1 + x 2 ) − ∫ 2 l n ( x + 1 + x 2 ) d ( 1 + x 2 ) = x l n 2 ( x + 1 + x 2 ) − 2 1 + x 2 l n ( x + 1 + x 2 ) + 2 x + C ( 34 ) 设 u = 1 x ，则 x = 1 u ， d x = − 1 u 2 d u ，得 ∫ l n x ( 1 + x 2 ) 3 2 d x = − ∫ l n 1 u ( 1 + 1 u 2 ) 3 2 ⋅ 1 u 2 d u = ∫ u l n u ( 1 + u 2 ) 3 2 d u = − ∫ l n u d ( ( 1 + u 2 ) − 1 2 ) ， 设 s = l n u ， d v = d ( ( 1 + u 2 ) − 1 2 ) ，则 d s = 1 u d u ， v = 1 1 + u 2 ，得 − ∫ l n u d ( ( 1 + u 2 ) − 1 2 ) = − l n u 1 + u 2 + ∫ 1 u 1 + u 2 d u = x l n x 1 + x 2 − ∫ 1 1 + x 2 d x = x l n x 1 + x 2 − l n ( x + 1 + x 2 ) + C ( 35 ) ∫ 1 − x 2 a r c s i n x d x = ∫ a r c s i n x 1 1 − x 2 d x − ∫ a r c s i n x x 2 1 − x 2 d x ， 设 u = a r c s i n x ， d v = 1 1 − x 2 d x ，则 d u = 1 1 − x 2 d x ， v = a r c s i n x ，得 ∫ a r c s i n x 1 1 − x 2 d x = ( a r c s i n x ) 2 − ∫ a r c s i n x 1 1 − x 2 d x ，得 ∫ a r c s i n x 1 1 − x 2 d x = 1 2 ( a r c s i n x ) 2 设 u = a r c s i n x ， d v = x 2 1 − x 2 d x ，则 d u = 1 1 − x 2 d x ， v = − x 2 + 1 2 a r c s i n x ，得 ∫ a r c s i n x x 2 1 − x 2 d x = 1 2 ( a r c s i n x ) 2 − 1 2 x 1 − x 2 a r c s i n x − 1 2 ∫ ( a r c s i n x − x 1 − x 2 ) 1 1 − x 2 d x = 1 2 ( a r c s i n x ) 2 − 1 2 x 1 − x 2 a r c s i n x − 1 2 ∫ a r c s i n x 1 1 − x 2 d x + 1 4 x 2 ，原式得 ∫ a r c s i n x 1 1 − x 2 d x − ∫ a r c s i n x x 2 1 − x 2 d x = 1 2 ∫ a r c s i n x 1 1 − x 2 d x + 1 2 x 1 − x 2 a r c s i n x − 1 4 x 2 = 1 4 ( a r c s i n x ) 2 + 1 2 x 1 − x 2 a r c s i n x − 1 4 x 2 + C ( 36 ) 设 x = c o s u ( 0 < u < π ) ，则 1 − x 2 = s i n u ， d x = − s i n u d u ，得 ∫ x 3 a r c c o s x 1 − x 2 d x = − ∫ u c o s 3 u d u = − ∫ u d ( s i n u − 1 3 s i n 3 u ) = − u ( s i n u − 1 3 s i n 3 u ) + ∫ ( s i n u − 1 3 s i n 3 u ) d u = − u ( s i n u − 1 3 s i n 3 u ) − 1 3 ∫ ( 2 + c o s 2 u ) d ( c o s u ) = − u ( s i n u − 1 3 s i n 3 u ) − 2 3 c o s u − 1 9 c o s 3 u + C = − 1 3 1 − x 2 ( 2 + x 2 ) a r c c o s x − 1 9 x ( 6 + x 2 ) + C ( 37 ) ∫ c o t x 1 + s i n x d x = ∫ c o s x s i n x 1 + s i n x d x = ∫ 1 − s i n 2 x s i n x ( 1 + s i n x ) d x ， 设 u = s i n x ，则 x = a r c s i n u ， d x = 1 1 − u 2 d u ，得 ∫ 1 − s i n 2 x s i n x ( 1 + s i n x ) d x = ∫ 1 − u 2 u ( 1 + u ) ⋅ 1 1 − u 2 d u = ∫ 1 u ( 1 + u ) d u = ∫ ( 1 u − 1 1 + u ) d u = l n ∣ s i n x 1 + s i n x ∣ + C ( 38 ) ∫ d x s i n 3 x c o s x = − ∫ c o t x s e c 2 x d ( c o t x ) ，设 u = c o t x ，则 x = a r c c o t u ， d x = − 1 1 + u 2 d u ，得 − ∫ c o t x s e c 2 x d ( c o t x ) = − ∫ u ( 1 + 1 u 2 ) d u = − 1 2 u 2 − l n ∣ u ∣ + C = − 1 2 c o t 2 x − l n ∣ c o t x ∣ + C ( 39 ) 设 u = c o s x ，则 x = a r c c o s u ， d x = − 1 1 − u 2 d u ，得 ∫ d x ( 2 + c o s x ) s i n x = ∫ 1 ( u + 2 ) ( u 2 − 1 ) d u = ∫ [ 1 6 ( u − 1 ) − 1 2 ( u + 1 ) + 1 3 ( u + 2 ) ] d u = 1 6 l n ∣ u − 1 ∣ − 1 2 l n ∣ u + 1 ∣ + 1 3 l n ∣ u + 2 ∣ + C = 1 6 l n ∣ 1 − c o s x ∣ − 1 2 l n ∣ c o s x + 1 ∣ + 1 3 l n ∣ c o s x + 2 ∣ + C ( 40 ) ∫ s i n x c o s x s i n x + c o s x d x = 1 2 ∫ ( s i n x + c o s x ) 2 − 1 s i n x + c o s x d x = 1 2 ∫ ( s i n x + c o s x ) d x − 1 2 ∫ 1 s i n x + c o s x d x = 1 2 ∫ ( s i n x − c o s x ) − 1 2 ∫ 1 s i n x + c o s x d x ， 令 u = t a n x 2 ，则 s i n x = 2 u 1 + u 2 ， c o s x = 1 − u 2 1 + u 2 ， d x = 1 1 + u 2 d u ，得 ∫ 1 s i n x + c o s x d x = ∫ 2 2 u + 1 − u 2 d u = − ∫ 2 ( u − 1 ) 2 − 2 d u = − 2 2 ∫ 1 u − 1 − 2 d u + 2 2 ∫ 1 u − 1 + 2 d u = 2 2 l n ∣ u − 1 + 2 u − 1 − 2 ∣ + C 代入原式得 ∫ s i n x c o s x s i n x + c o s x d x = 1 2 ( s i n x − c o s x ) − 2 4 l n ∣ t a n x 2 − 1 + 2 t a n x 2 − 1 − 2 ∣ + C \begin{aligned} &\ \ (20)\ \int \frac{sin^2\ x}{cos^3\ x}dx=\int cos^{-3}\ xsin^2\ xdx，根据积分表（十一）公式99和98，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int cos^{-3}\ xsin^2\ xdx=cos^{-2}\ xsin\ x-\int cos^{-3}\ xdx=cos^{-2}\ xsin\ x-\frac{1}{2}cos^{-2}\ xsin\ x-\frac{1}{2}\int \frac{1}{cos\ x}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}cos^{-2}\ xsin\ x-\frac{1}{2}ln\ |sec\ x+tan\ x|+C=\frac{1}{2}sec\ xtan\ x-\frac{1}{2}ln\ |sec\ x+tan\ x|+C\\\\ &\ \ (21)\ 设u=arctan\sqrt{x}，则x=tan^2\ u，dx=2tan\ usec^2\ udu，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arctan\ \sqrt{x}dx=2\int utan\ usec^2\ udu=2\int usec\ ud(sec\ u)，设s=sec\ u，ds=d(sec\ u)，则u=arccos\ \frac{1}{s}，\\\\ &\ \ \ \ \ \ \ \ \ \ 得2\int usec\ ud(sec\ u)=2\int arccos\ \frac{1}{s}\cdot sds，令u=arccos\ \frac{1}{s}，dv=sds，则du=-\frac{s}{\sqrt{s^2-1}}ds，v=\frac{1}{2}s^2，得\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int arccos\ \frac{1}{s}\cdot sds=s^2arccos\ \frac{1}{s}+\int \frac{s^3}{\sqrt{s^2-1}}ds，根据积分表（八）公式66得，\\\\ &\ \ \ \ \ \ \ \ \ \ s^2arccos\ \frac{1}{s}+\int \frac{s^3}{\sqrt{s^2-1}}ds=s^2arccos\ \frac{1}{s}-\sqrt{s^2-1}+C=(1+x)arctan\sqrt{x}-\sqrt{x}+C\\\\ &\ \ (22)\ \int \frac{\sqrt{1+cos\ x}}{sin\ x}dx=\int \frac{\sqrt{2}\left|cos\frac{x}{2}\right|}{2sin\frac{x}{2}cos\frac{x}{2}}dx=\pm\sqrt{2}\int csc\frac{x}{2}d\left(\frac{x}{2}\right)=\pm\sqrt{2}ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|+C\\\\ &\ \ \ \ \ \ \ \ \ \ 当cos\frac{x}{2} \gt 0时，ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|=ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)\\\\ &\ \ \ \ \ \ \ \ \ \ 当cos\frac{x}{2} \lt 0时，ln\left|csc\frac{x}{2}-cot\frac{x}{2}\right|=ln\left(\left|csc\frac{x}{2}\right|+\left|cot\frac{x}{2}\right|\right)=-ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)\\\\ &\ \ \ \ \ \ \ \ \ \ 因此，\int \frac{\sqrt{1+cos\ x}}{sin\ x}dx=\sqrt{2}ln\left(\left|csc\frac{x}{2}\right|-\left|cot\frac{x}{2}\right|\right)+C\\\\ &\ \ (23)\ \int \frac{x^3}{(1+x^8)^2}dx=\int \frac{x^4}{x(1+(x^4)^2)^2}dx，设u=x^4，x=\sqrt[4]{x}，dx=-\frac{1}{4\sqrt[4]{u^3}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^3}{(1+x^8)^2}dx=\int \frac{u}{\sqrt[4]{u}(1+u^2)^2}\cdot \frac{1}{4\sqrt[4]{u^3}}du=\frac{1}{4}\int \frac{1}{(1+u^2)^2}du，根据积分表（四）公式28，得\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}\left(\frac{u}{2(u^2+1)}+\frac{1}{2}\int \frac{1}{1+u^2}du\right)=\frac{u}{8(u^2+1)}+\frac{1}{8}arctan\ u+C=\frac{x^4}{8(x^8+1)}+\frac{1}{8}arctan\ x^4+C\\\\ &\ \ (24)\ 设u=x^4，x=\sqrt[4]{u}，则dx=\frac{1}{4\sqrt[4]{u^3}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^{11}}{x^8+3x^4+2}dx=\int \frac{\sqrt[4]{u^{11}}}{u^2+3u+2}\cdot \frac{1}{4\sqrt[4]{u^3}}du=\frac{1}{4}\int \frac{u^2}{u^2+3u+2}du=\frac{1}{4}\int \left(1+\frac{1}{u+1}-\frac{4}{u+2}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}u+\frac{1}{4}ln\ |1+u|-ln\ |2+u|+C=\frac{1}{4}x^4+ln\ \frac{\sqrt[4]{1+x^4}}{2+x^4}+C\\\\ &\ \ (25)\ \int \frac{dx}{16-x^4}=\frac{1}{8}\int \left(\frac{1}{4+x^2}+\frac{1}{4-x^2}\right)dx=\frac{1}{8}\int \frac{1}{4+x^2}dx+\frac{1}{32}\int \left(\frac{1}{2+x}+\frac{1}{2-x}\right)dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{16}arctan\frac{x}{2}+\frac{1}{32}ln\ \left|\frac{2+x}{2-x}\right|+C\\\\ &\ \ (26)\ \int \frac{sin\ x}{1+sin\ x}dx=\int \frac{\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}{\frac{\left(tan\frac{x}{2}+1\right)^2}{1+tan^2\frac{x}{2}}}dx=\int \frac{2tan\frac{x}{2}}{\left(tan\frac{x}{2}+1\right)^2}dx，设u=tan\frac{x}{2}，则x=2arctan\ u，dx=\frac{2}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{2tan\frac{x}{2}}{\left(tan\frac{x}{2}+1\right)^2}dx=4\int \frac{u}{(u+1)^2}\cdot \frac{1}{1+u^2}du=2\int \left(\frac{1}{1+u^2}-\frac{1}{(1+u)^2}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int \frac{1}{1+u^2}du-2\int \frac{1}{(u+1)^2}du=2arctan\ u+\frac{2}{1+u}+C=x+\frac{2}{1+tan\frac{x}{2}}+C\\\\ &\ \ (27)\ \int \frac{x+sin\ x}{1+cos\ x}dx=\int \frac{x+2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}dx=2\int \frac{x}{2}sec^2\frac{x}{2}d\left(\frac{x}{2}\right)+\int tan\frac{x}{2}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=\frac{x}{2}，dv=sec^2\frac{x}{2}d\left(\frac{x}{2}\right)，则du=\frac{1}{2}dx，v=tan\frac{x}{2}，得\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int \frac{x}{2}sec^2\frac{x}{2}d\left(\frac{x}{2}\right)+\int tan\frac{x}{2}dx=xtan\frac{x}{2}-\int tan\frac{x}{2}dx+\int tan\frac{x}{2}dx+C=xtan\frac{x}{2}+C\\\\ &\ \ (28)\ \int e^{sin\ x}\frac{xcos^3\ x-sin\ x}{cos^2\ x}dx=\int xe^{sin\ x}cos\ xdx=\int e^{sin\ x}tan\ xsec\ xdx=\int xd(e^{sin\ x})-\int e^{sin\ x}d(sec\ x)\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=x，dv=d(e^{sin\ x})，则du=dx，v=e^{sin\ x}，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int xd(e^{sin\ x})=xe^{sin\ x}-\int e^{sin\ x}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=e^{sin\ x}，dv=d(sec\ x)，则du=e^{sin\ x}cos\ xdx，v=sec\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int e^{sin\ x}d(sec\ x)=sec\ xe^{sin\ x}-\int e^{sin\ x}dx，因此\\\\ &\ \ \ \ \ \ \ \ \ \ \int xd(e^{sin\ x})-\int e^{sin\ x}d(sec\ x)=xe^{sin\ x}-\int e^{sin\ x}dx-sec\ xe^{sin\ x}+\int e^{sin\ x}dx=(x-sec\ x)e^{sin\ x}+C\\\\ &\ (29)\ 设u=\sqrt[6]{x}，则x=u^6，dx=6u^5du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{\sqrt[3]{x}}{x(\sqrt{x}+\sqrt[3]{x})}dx=\int \frac{u^2}{u^6(u^3+u^2)}\cdot 6u^5du=6\int \frac{1}{u(u+1)}du=6\int \left(\frac{1}{u}-\frac{1}{u+1}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ 6ln\ \left|\frac{u}{u+1}\right|+C=6ln\frac{\sqrt[6]{x}}{\sqrt[6]{x}+1}+C=ln\frac{x}{(\sqrt[6]{x}+1)^6}+C\\\\ &\ \ (30)\ 设u=e^x，则x=ln\ u，dx=\frac{1}{u}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{(1+e^x)^2}=\int \frac{1}{u(1+u)^2}du=\int \left(\frac{1}{u}-\frac{1}{1+u}-\frac{1}{(1+u)^2}\right)du=ln\ u-ln(1+u)+\frac{1}{1+u}+C=\\\\ &\ \ \ \ \ \ \ \ \ \ x-ln(1+e^x)+\frac{1}{1+e^x}+C\\\\ &\ \ (31)\ \int \frac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}dx=\int \frac{e^x+e^{-x}}{e^{2x}-1+e^{-2x}}dx=\int \frac{1}{(e^x-e^{-x})^2+1}d(e^x-e^{-x})=arctan(e^x-e^{-x})+C\\\\ &\ \ (32)\ 设u=e^x，则x=ln\ u，dx=\frac{1}{u}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{xe^x}{(e^x+1)^2}dx=\int \frac{ln\ u}{(1+u)^2}du，令s=ln\ u，dv=\frac{1}{(1+u)^2}du，则ds=\frac{1}{u}du，v=-\frac{1}{1+u}，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{ln\ u}{(1+u)^2}du=-\frac{ln\ u}{1+u}+\int \frac{1}{u(1+u)}du=-\frac{ln\ u}{1+u}+\int \frac{1}{u}du-\int \frac{1}{1+u}du=\\\\ &\ \ \ \ \ \ \ \ \ \ -\frac{ln\ u}{1+u}+ln\ u-ln(1+u)+C=-\frac{x}{1+e^x}+x-ln(1+e^x)+C\\\\ &\ \ (33)\ 设u=ln^2(x+\sqrt{1+x^2})，dv=dx，则u=\frac{2ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx，v=x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int ln^2(x+\sqrt{1+x^2})dx=xln^2(x+\sqrt{1+x^2})-\int \frac{2xln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ xln^2(x+\sqrt{1+x^2})-\int 2ln(x+\sqrt{1+x^2})d(\sqrt{1+x^2})=\\\\ &\ \ \ \ \ \ \ \ \ \ xln^2(x+\sqrt{1+x^2})-2\sqrt{1+x^2}ln(x+\sqrt{1+x^2})+2x+C\\\\ &\ \ (34)\ 设u=\frac{1}{x}，则x=\frac{1}{u}，dx=-\frac{1}{u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{ln\ x}{(1+x^2)^{\frac{3}{2}}}dx=-\int \frac{ln\frac{1}{u}}{\left(1+\frac{1}{u^2}\right)^{\frac{3}{2}}}\cdot \frac{1}{u^2}du=\int \frac{uln\ u}{(1+u^2)^{\frac{3}{2}}}du=-\int ln\ ud((1+u^2)^{-\frac{1}{2}})，\\\\ &\ \ \ \ \ \ \ \ \ \ 设s=ln\ u，dv=d((1+u^2)^{-\frac{1}{2}})，则ds=\frac{1}{u}du，v=\frac{1}{1+u^2}，得\\\\ &\ \ \ \ \ \ \ \ \ \ -\int ln\ ud((1+u^2)^{-\frac{1}{2}})=-\frac{ln\ u}{\sqrt{1+u^2}}+\int \frac{1}{u\sqrt{1+u^2}}du=\frac{xln\ x}{\sqrt{1+x^2}}-\int \frac{1}{\sqrt{1+x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{xln\ x}{\sqrt{1+x^2}}-ln(x+\sqrt{1+x^2})+C\\\\ &\ \ (35)\ \int \sqrt{1-x^2}arcsin\ xdx=\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx-\int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=arcsin\ x，dv=\frac{1}{\sqrt{1-x^2}}dx，则du=\frac{1}{\sqrt{1-x^2}}dx，v=arcsin\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx=(arcsin\ x)^2-\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx，得\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx=\frac{1}{2}(arcsin\ x)^2\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=arcsin\ x，dv=\frac{x^2}{\sqrt{1-x^2}}dx，则du=\frac{1}{\sqrt{1-x^2}}dx，v=-\frac{x}{2}+\frac{1}{2}arcsin\ x，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx=\frac{1}{2}(arcsin\ x)^2-\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{2}\int (arcsin\ x-x\sqrt{1-x^2})\frac{1}{\sqrt{1-x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}(arcsin\ x)^2-\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{2}\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx+\frac{1}{4}x^2，原式得\\\\ &\ \ \ \ \ \ \ \ \ \ \int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx-\int arcsin\ x\frac{x^2}{\sqrt{1-x^2}}dx=\frac{1}{2}\int arcsin\ x\frac{1}{\sqrt{1-x^2}}dx+\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{4}x^2=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}(arcsin\ x)^2+\frac{1}{2}x\sqrt{1-x^2}arcsin\ x-\frac{1}{4}x^2+C\\\\ &\ \ (36)\ 设x=cos\ u\ (0 \lt u \lt \pi)，则\sqrt{1-x^2}=sin\ u，dx=-sin\ udu，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{x^3arccos\ x}{\sqrt{1-x^2}}dx=-\int ucos^3\ udu=-\int ud\left(sin\ u-\frac{1}{3}sin^3\ u\right)=\\\\ &\ \ \ \ \ \ \ \ \ \ -u\left(sin\ u-\frac{1}{3}sin^3\ u\right)+\int \left(sin\ u-\frac{1}{3}sin^3\ u\right)du=-u\left(sin\ u-\frac{1}{3}sin^3\ u\right)-\frac{1}{3}\int (2+cos^2\ u)d(cos\ u)=\\\\ &\ \ \ \ \ \ \ \ \ \ -u\left(sin\ u-\frac{1}{3}sin^3\ u\right)-\frac{2}{3}cos\ u-\frac{1}{9}cos^3\ u+C=-\frac{1}{3}\sqrt{1-x^2}(2+x^2)arccos\ x-\frac{1}{9}x(6+x^2)+C\\\\ &\ \ (37)\ \int \frac{cot\ x}{1+sin\ x}dx=\int \frac{\frac{cos\ x}{sin\ x}}{1+sin\ x}dx=\int \frac{\sqrt{1-sin^2\ x}}{sin\ x(1+sin\ x)}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 设u=sin\ x，则x=arcsin\ u，dx=\frac{1}{\sqrt{1-u^2}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{\sqrt{1-sin^2\ x}}{sin\ x(1+sin\ x)}dx=\int \frac{\sqrt{1-u^2}}{u(1+u)}\cdot \frac{1}{\sqrt{1-u^2}}du=\int \frac{1}{u(1+u)}du=\int \left(\frac{1}{u}-\frac{1}{1+u}\right)du=\\\\ &\ \ \ \ \ \ \ \ \ \ ln\ \left|\frac{sin\ x}{1+sin\ x}\right|+C\\\\ &\ \ (38)\ \int \frac{dx}{sin^3\ xcos\ x}=-\int cot\ xsec^2\ xd(cot\ x)，设u=cot\ x，则x=arccot\ u，dx=-\frac{1}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ -\int cot\ xsec^2\ xd(cot\ x)=-\int u\left(1+\frac{1}{u^2}\right)du=-\frac{1}{2}u^2-ln\ |u|+C=-\frac{1}{2}cot^2\ x-ln\ |cot\ x|+C\\\\ &\ \ (39)\ 设u=cos\ x，则x=arccos\ u，dx=-\frac{1}{\sqrt{1-u^2}}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{(2+cos\ x)sin\ x}=\int \frac{1}{(u+2)(u^2-1)}du=\int \left[\frac{1}{6(u-1)}-\frac{1}{2(u+1)}+\frac{1}{3(u+2)}\right]du=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{6}ln\ |u-1|-\frac{1}{2}ln\ |u+1|+\frac{1}{3}ln\ |u+2|+C=\frac{1}{6}ln\ |1-cos\ x|-\frac{1}{2}ln\ |cos\ x+1|+\frac{1}{3}ln\ |cos\ x+2|+C\\\\ &\ \ (40)\ \int \frac{sin\ xcos\ x}{sin\ x+cos\ x}dx=\frac{1}{2}\int \frac{(sin\ x+cos\ x)^2-1}{sin\ x+cos\ x}dx=\frac{1}{2}\int (sin\ x+cos\ x)dx-\frac{1}{2}\int \frac{1}{sin\ x+cos\ x}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int (sin\ x-cos\ x)-\frac{1}{2}\int \frac{1}{sin\ x+cos\ x}dx，\\\\ &\ \ \ \ \ \ \ \ \ \ 令u=tan\frac{x}{2}，则sin\ x=\frac{2u}{1+u^2}，cos\ x=\frac{1-u^2}{1+u^2}，dx=\frac{1}{1+u^2}du，得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{1}{sin\ x+cos\ x}dx=\int \frac{2}{2u+1-u^2}du=-\int \frac{2}{(u-1)^2-2}du=\\\\ &\ \ \ \ \ \ \ \ \ \ -\frac{\sqrt{2}}{2}\int \frac{1}{u-1-\sqrt{2}}du+\frac{\sqrt{2}}{2}\int \frac{1}{u-1+\sqrt{2}}du=\frac{\sqrt{2}}{2}ln\ \left|\frac{u-1+\sqrt{2}}{u-1-\sqrt{2}}\right|+C\\\\ &\ \ \ \ \ \ \ \ \ \ 代入原式得\int \frac{sin\ xcos\ x}{sin\ x+cos\ x}dx=\frac{1}{2}(sin\ x-cos\ x)-\frac{\sqrt{2}}{4}ln\ \left|\frac{tan\frac{x}{2}-1+\sqrt{2}}{tan\frac{x}{2}-1-\sqrt{2}}\right|+C & \end{aligned} (20) ∫cos3 xsin2 xdx=∫cos−3 xsin2 xdx，根据积分表（十一）公式99和98，得 ∫cos−3 xsin2 xdx=cos−2 xsin x−∫cos−3 xdx=cos−2 xsin x−21cos−2 xsin x−21∫cos x1dx= 21cos−2 xsin x−21ln ∣sec x+tan x∣+C=21sec xtan x−21ln ∣sec x+tan x∣+C (21) 设u=arctanx，则x=tan2 u，dx=2tan usec2 udu，得 ∫arctan xdx=2∫utan usec2 udu=2∫usec ud(sec u)，设s=sec u，ds=d(sec u)，则u=arccos s1， 得2∫usec ud(sec u)=2∫arccos s1⋅sds，令u=arccos s1，dv=sds，则du=−s2−1sds，v=21s2，得 2∫arccos s1⋅sds=s2arccos s1+∫s2−1s3ds，根据积分表（八）公式66得， s2arccos s1+∫s2−1s3ds=s2arccos s1−s2−1+C=(1+x)arctanx−x+C (22) ∫sin x1+cos xdx=∫2sin2xcos2x2∣ ∣cos2x∣ ∣dx=±2∫csc2xd(2x)=±2ln∣ ∣csc2x−cot2x∣ ∣+C 当cos2x>0时，ln∣ ∣csc2x−cot2x∣ ∣=ln(∣ ∣csc2x∣ ∣−∣ ∣cot2x∣ ∣) 当cos2x<0时，ln∣ ∣csc2x−cot2x∣ ∣=ln(∣ ∣csc2x∣ ∣+∣ ∣cot2x∣ ∣)=−ln(∣ ∣csc2x∣ ∣−∣ ∣cot2x∣ ∣) 因此，∫sin x1+cos xdx=2ln(∣ ∣csc2x∣ ∣−∣ ∣cot2x∣ ∣)+C (23) ∫(1+x8)2x3dx=∫x(1+(x4)2)2x4dx，设u=x4，x=4x，dx=−44u31du，得 ∫(1+x8)2x3dx=∫4u(1+u2)2u⋅44u31du=41∫(1+u2)21du，根据积分表（四）公式28，得 41(2(u2+1)u+21∫1+u21du)=8(u2+1)u+81arctan u+C=8(x8+1)x4+81arctan x4+C (24) 设u=x4，x=4u，则dx=44u31du，得 ∫x8+3x4+2x11dx=∫u2+3u+24u11⋅44u31du=41∫u2+3u+2u2du=41∫(1+u+11−u+24)du= 41u+41ln ∣1+u∣−ln ∣2+u∣+C=41x4+ln 2+x441+x4+C (25) ∫16−x4dx=81∫(4+x21+4−x21)dx=81∫4+x21dx+321∫(2+x1+2−x1)dx= 161arctan2x+321ln ∣ ∣2−x2+x∣ ∣+C (26) ∫1+sin xsin xdx=∫1+tan22x(tan2x+1)21+tan22x2tan2xdx=∫(tan2x+1)22tan2xdx，设u=tan2x，则x=2arctan u，dx=1+u22du，得 ∫(tan2x+1)22tan2xdx=4∫(u+1)2u⋅1+u21du=2∫(1+u21−(1+u)21)du= 2∫1+u21du−2∫(u+1)21du=2arctan u+1+u2+C=x+1+tan2x2+C (27) ∫1+cos xx+sin xdx=∫2cos22xx+2sin2xcos2xdx=2∫2xsec22xd(2x)+∫tan2xdx， 设u=2x，dv=sec22xd(2x)，则du=21dx，v=tan2x，得 2∫2xsec22xd(2x)+∫tan2xdx=xtan2x−∫tan2xdx+∫tan2xdx+C=xtan2x+C (28) ∫esin xcos2 xxcos3 x−sin xdx=∫xesin xcos xdx=∫esin xtan xsec xdx=∫xd(esin x)−∫esin xd(sec x) 设u=x，dv=d(esin x)，则du=dx，v=esin x，得 ∫xd(esin x)=xesin x−∫esin xdx， 设u=esin x，dv=d(sec x)，则du=esin xcos xdx，v=sec x，得 ∫esin xd(sec x)=sec xesin x−∫esin xdx，因此 ∫xd(esin x)−∫esin xd(sec x)=xesin x−∫esin xdx−sec xesin x+∫esin xdx=(x−sec x)esin x+C (29) 设u=6x，则x=u6，dx=6u5du，得 ∫x(x+3x)3xdx=∫u6(u3+u2)u2⋅6u5du=6∫u(u+1)1du=6∫(u1−u+11)du= 6ln ∣ ∣u+1u∣ ∣+C=6ln6x+16x+C=ln(6x+1)6x+C (30) 设u=ex，则x=ln u，dx=u1du，得 ∫(1+ex)2dx=∫u(1+u)21du=∫(u1−1+u1−(1+u)21)du=ln u−ln(1+u)+1+u1+C= x−ln(1+ex)+1+ex1+C (31) ∫e4x−e2x+1e3x+exdx=∫e2x−1+e−2xex+e−xdx=∫(ex−e−x)2+11d(ex−e−x)=arctan(ex−e−x)+C (32) 设u=ex，则x=ln u，dx=u1du，得 ∫(ex+1)2xexdx=∫(1+u)2ln udu，令s=ln u，dv=(1+u)21du，则ds=u1du，v=−1+u1，得 ∫(1+u)2ln udu=−1+uln u+∫u(1+u)1du=−1+uln u+∫u1du−∫1+u1du= −1+uln u+ln u−ln(1+u)+C=−1+exx+x−ln(1+ex)+C (33) 设u=ln2(x+1+x2)，dv=dx，则u=1+x22ln(x+1+x2)dx，v=x，得 ∫ln2(x+1+x2)dx=xln2(x+1+x2)−∫1+x22xln(x+1+x2)dx= xln2(x+1+x2)−∫2ln(x+1+x2)d(1+x2)= xln2(x+1+x2)−21+x2ln(x+1+x2)+2x+C (34) 设u=x1，则x=u1，dx=−u21du，得 ∫(1+x2)23ln xdx=−∫(1+u21)23lnu1⋅u21du=∫(1+u2)23uln udu=−∫ln ud((1+u2)−21)， 设s=ln u，dv=d((1+u2)−21)，则ds=u1du，v=1+u21，得 −∫ln ud((1+u2)−21)=−1+u2ln u+∫u1+u21du=1+x2xln x−∫1+x21dx= 1+x2xln x−ln(x+1+x2)+C (35) ∫1−x2arcsin xdx=∫arcsin x1−x21dx−∫arcsin x1−x2x2dx， 设u=arcsin x，dv=1−x21dx，则du=1−x21dx，v=arcsin x，得 ∫arcsin x1−x21dx=(arcsin x)2−∫arcsin x1−x21dx，得∫arcsin x1−x21dx=21(arcsin x)2 设u=arcsin x，dv=1−x2x2dx，则du=1−x21dx，v=−2x+21arcsin x，得 ∫arcsin x1−x2x2dx=21(arcsin x)2−21x1−x2arcsin x−21∫(arcsin x−x1−x2)1−x21dx= 21(arcsin x)2−21x1−x2arcsin x−21∫arcsin x1−x21dx+41x2，原式得 ∫arcsin x1−x21dx−∫arcsin x1−x2x2dx=21∫arcsin x1−x21dx+21x1−x2arcsin x−41x2= 41(arcsin x)2+21x1−x2arcsin x−41x2+C (36) 设x=cos u (0<u<π)，则1−x2=sin u，dx=−sin udu，得 ∫1−x2x3arccos xdx=−∫ucos3 udu=−∫ud(sin u−31sin3 u)= −u(sin u−31sin3 u)+∫(sin u−31sin3 u)du=−u(sin u−31sin3 u)−31∫(2+cos2 u)d(cos u)= −u(sin u−31sin3 u)−32cos u−91cos3 u+C=−311−x2(2+x2)arccos x−91x(6+x2)+C (37) ∫1+sin xcot xdx=∫1+sin xsin xcos xdx=∫sin x(1+sin x)1−sin2 xdx， 设u=sin x，则x=arcsin u，dx=1−u21du，得 ∫sin x(1+sin x)1−sin2 xdx=∫u(1+u)1−u2⋅1−u21du=∫u(1+u)1du=∫(u1−1+u1)du= ln ∣ ∣1+sin xsin x∣ ∣+C (38) ∫sin3 xcos xdx=−∫cot xsec2 xd(cot x)，设u=cot x，则x=arccot u，dx=−1+u21du，得 −∫cot xsec2 xd(cot x)=−∫u(1+u21)du=−21u2−ln ∣u∣+C=−21cot2 x−ln ∣cot x∣+C (39) 设u=cos x，则x=arccos u，dx=−1−u21du，得 ∫(2+cos x)sin xdx=∫(u+2)(u2−1)1du=∫[6(u−1)1−2(u+1)1+3(u+2)1]du= 61ln ∣u−1∣−21ln ∣u+1∣+31ln ∣u+2∣+C=61ln ∣1−cos x∣−21ln ∣cos x+1∣+31ln ∣cos x+2∣+C (40) ∫sin x+cos xsin xcos xdx=21∫sin x+cos x(sin x+cos x)2−1dx=21∫(sin x+cos x)dx−21∫sin x+cos x1dx= 21∫(sin x−cos x)−21∫sin x+cos x1dx， 令u=tan2x，则sin x=1+u22u，cos x=1+u21−u2，dx=1+u21du，得 ∫sin x+cos x1dx=∫2u+1−u22du=−∫(u−1)2−22du= −22∫u−1−21du+22∫u−1+21du=22ln ∣ ∣u−1−2u−1+2∣ ∣+C 代入原式得∫sin x+cos xsin xcos xdx=21(sin x−cos x)−42ln ∣ ∣tan2x−1−2tan2x−1+2∣ ∣+C