int(1)/(cos2x+sin^(2)x)dx 01:10 Evaluate: (i) int(tanx+cotx)^2\ dx (ii) int(1-cos2x)/(1+cos2x)\ dx 03:57 intsinxsqrt(1+cos2x)\ dx 01:45 intsinx\ sqrt(1-cos2x)\ dx 02:14 Evaluate: (i) intsqrt((1-cos2x)/(1+cos2x))\ \ dx (ii) intsqrt((1+cosx)... 03...
int sin^(-1)(cos2x)*dx View Solution Evaluate: (i)∫11+cos2xdx (ii)∫11−cos2xdx View Solution Evaluate :∫tan−1{√1−cos2x1+cos2x}dx. View Solution (i)∫1(1+cos2x)dx, (ii)∫1(1−cos2x)dx View Solution Evaluate the following integrals: (i)∫(1−cos2x1+cos2x)d...
实际上并不麻烦;考察该不定积分, 利用恒等式化简:∫1+cos2xdx=2∫|cosx|dx此处去绝对值分...
$$ \int \cos 2 x d x = ( ) $$$ A . - \fra c { 1 } { 2 } \sin 2 x + c $$$ B . 2 s i n 2 x + c $$$ C . s i n 2 x + c $$$ D . s i n x c o s x + c $$ 相关知识点: 试题来源: 解析 2∫cos2xdx=∫cos2xd2x=sin2x+c ...
【解析】 $$ \frac{x}{\sqrt{1-x^{2}}}dx \\ =- \frac{1}{2}\int \frac{1}{\sqrt{1-x^{2}}}d(1-x^{2}) \\ =- \frac{1}{2}\int \frac{1}{\sqrt{t}}xt=- \frac{1}{2}\sqrt{1-x^{2}}+c(\in \in R) $$ 结果四 题目 【题目】求积分:∫_0^πx...
Evaluate: int (x-3)(5-x) dx 01:37 Evaluate: inte^(a x)cosb x\ dx 06:44 Evaluate: inte^(-x)cosx\ dx 03:20 Evaluate: inte^(2x)((1+sin2x)/(1+cos2x))dx 03:54 Evaluate: (i)intcos( b x+c)dx (ii) intsin(logx)dx 05:08 int1/(1+tanx)\ dx= (log)e(x+sinx)+C ...
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} \int^{\frac{\pi }{2} }_{-\frac{\pi }{2} } \sqrt{1+\cos 2x} \ dx\\ &\x...
解$$ \int \frac { x } { 1 + \cos 2 x } d x = \int \frac { x } { 2 \cos ^ { 2 } x } d x = \frac { 1 } { 2 } \int x d \tan \\ = \frac { 1 } { 2 } x \tan x - \frac { 1 } { 2 } \int \tan x d x \\ = \frac { 1 } { ...
解析 【解析】 $$ 1+ \cot ^{2}x= \frac{1}{\sin ^{2}x} $$ $$ \cot^{2}x= \frac{1}{\sin ^{2}x}-1 $$ $$ \int(\frac{1}{\sin ^{2}x}-1)dx \\ = \int \frac{dx}{\sin ^{2}x}- \int 1dx \\ =-10+x-x+c $$ ...