sinx+cosx Evaluate:∫sinx+cosx√1−sin2xdx View Solution Evaluate:∫sinx+cosx√1−sin2xdx View Solution Evaluate:∫sinx+cosx√1−sin2xdx View Solution Write a value of∫sinx−cosx√1+sin2xdx View Solution Write a value of∫sinx+cosx√1+sin2xdx ...
左端=\int_{0}^{\frac{\pi}{2} } {\frac{sinx}{sinx+cox} }\,{\rm dx}=\int_{\frac{\pi}{2} }^{0} {\frac{sin(\frac{\pi}{2}-t )}{sin(\frac{\pi}{2}-t )+cos(\frac{\pi}{2} -t)} }\,{\rm (-dt)}=\int_{0}^{\frac{\pi}{2} } {\frac{cost}{sint...
百度试题 结果1 题目 \(\int_ {0}^{\dfrac {\pi} {2}}\sqrt{1-sin2x}dx=\int_ {0}^{\dfrac {\pi} {2}}(sinx-cosx)dx=(-cosx-sinx)|_0^\dfrac {\pi} {2}=0\) 相关知识点: 试题来源: 解析 错误 反馈 收藏
结果1 题目 \(\int_ {0}^{\dfrac {\pi} {2}}\dfrac{sinx}{1+cosx}dx=-\int_ {0}^{\dfrac {\pi} {2}}\dfrac{1}{1+cosx}dcosx=(令cosx=t) -\int_ {0}^{1}\dfrac{1}{1+t}dt=-ln|t+1||_1^0=-ln2\) 相关知识点: 试题来源: 解析 false 反馈 收藏 ...
\int {\dfrac{lntanx}{cosxsinx} }\,{\rm dx} 相关知识点: 试题来源: 解析 原式=\int {lntanx}\,{\rm d(lntanx)} =\dfrac{1}{2}(lntanx)^2+C 本题考查不定积分的求解,对于特殊的不定积分进行记忆,可以使求解变得更为简便。反馈 收藏
If f(x)=cosxcos2xcos4xcos8xcos16x then f'(π4) equals View Solution ∫dxsinxcosx+2cos2x= View Solution Recommended Questions The value of int (sinx cosx cos2x cos4x cos8x cos16x) dx is equal to 02:28 The value of int( sin )x cos x cos2x cos4x cos8x cos16x)dx is int(sin...
\int {\cos x-\sin x}\,{\rm dx}=( ) A. -sinx+cosx+C B. sinx-cosx+C C. -sinx-cosx+C D
左端=\int_{0}^{\dfrac{\pi}{2} } {\dfrac{sinx}{sinx+cox} }\,{\rm dx}=\int_{\dfrac{\pi}{2} }^{0} {\dfrac{sin(\dfrac{\pi}{2}-t )}{sin(\dfrac{\pi}{2}-t )+cos(\dfrac{\pi}{2} -t)} }\,{\rm (-dt)}=\int_{0}^{\dfrac{\pi}{2} } {\dfrac{cost}...
【解析】∫(√(x^4+x^4+2))/(x^3)dx=∫ rac(√((x^2+x^(-2))^2)(x^3)dx=∫( rac(x^2+x^(-2) =∫(1/x+1/(x^3))dx=ln|x|-1/(4x^4)+C 结果二 题目 【题目】求下列不定积分: 答案 【解析】 2x-4x+41n(x+1)+C. 结果三 题目 求下列不定积分: 答...
StepsExamples simplifyintsin(x)−xcos(x) Solution intsin(x)−xcos(x) Examples x2−x−6=0 −x+3>2x+1 line(1,2),(3,1) ′ ∫ ∫ ∑ Connection error, please try again OK