A) \[\frac{\pi }{2}\left( \sqrt{2}+1 \right)\]
B) \[\pi \left( \sqrt{2}-1 \right)\]
C) \[2\pi \left( \sqrt{2}-1 \right)\]
D) \[\pi \sqrt{2}\]
Correct Answer: A
Solution :
Given integral is \[I=\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\frac{x}{1+\sin x}dx}\] As in the denominator there is\[1+\sin x\], we first rationalise the denominator and then do integration by parts So multiplying numerator and denominator by \[1-\sin x\] , we get \[\frac{x(1-\sin x)}{1-{{(\sin x)}^{2}}}=\frac{x(1-\sin x)}{{{(\cos x)}^{2}}}\] \[\Rightarrow =x(1-\sin x){{\sec }^{2}}x\] \[=x{{\sec }^{2}}x-x\sin x{{\sec }^{2}}x=x{{\sec }^{2}}x-x\tan x\sec x\] Now applying integration by parts to this \[\int_{{}}^{{}}{uvdx=u}\int_{{}}^{{}}{vdx-\int_{{}}^{{}}{\frac{du}{dx}\times }\int_{{}}^{{}}{vdx}}\] Therefore by applying the above formula we get \[I=\int_{{}}^{{}}{x{{\sec }^{2}}xdx-}\int_{{}}^{{}}{x\sec x\tan xdx}\] \[=\left[ x\tan x-\int_{{}}^{{}}{\frac{dx}{dx}\tan xdx} \right]-\left[ x\sec x-\int_{{}}^{{}}{\frac{dx}{dx}\sec xdx} \right]\] \[=[x\tan x-In|\sec x|]-[x\sec x-In|\sec x+\tan x|+c\] Now substituting the limits \[\frac{\pi }{4}\] and \[\frac{3\pi }{4}\] we get \[\begin{align} & I=\left\{ \left[ \frac{3\pi }{4}\tan \frac{3\pi }{4}-In\left| \frac{3\pi }{4} \right| \right]-\left[ \frac{3\pi }{4}\sec \frac{3\pi }{4}-In\left| \sec \frac{3\pi }{4}+\tan \frac{3\pi }{4} \right| \right] \right\}- \\ & \left\{ \left[ \frac{\pi }{4}\tan \frac{\pi }{4}-In\left| \frac{\pi }{4} \right| \right]-\left[ \frac{\pi }{4}\sec \frac{\pi }{4}-In|\sec \frac{\pi }{4}+\tan \frac{\pi }{4}| \right] \right\} \\ \end{align}\] \[=\frac{\pi }{2}(\sqrt{2}+1)\] We have, \[I=\int_{\frac{\pi }{4}\grave{\ }}^{\frac{3\pi }{4}}{\frac{x}{1+\sin x}dx}\] Multiply and divide LHS with \[(1-\sin x)\], we get \[I=\int_{\frac{\pi }{4}}^{\frac{3x}{4}}{\frac{x(1-\sin x)}{1-{{\sin }^{2}}x}dx}\] \[I=\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\frac{x(1-\sin x)}{{{\cos }^{2}}x}dx}\] \[I=\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{x({{\sec }^{2}}x-\sec x\tan x)dx}\] \[I=\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{x{{\sec }^{2}}x-}\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{x\sec x\tan xdx}\] Using integration by parts \[I=[(x\tan x)-(\int_{{}}^{{}}{1.\tan xdx)]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}+[(x\sec x)-(\int_{{}}^{{}}{1.\sec xdx)]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}}}\]\[I=[(x\tan x)-(log|secx|]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}+[(x\sec x)+(\log |\sec x+\tan x|]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\] \[I=\frac{\pi }{2}(\sqrt{2}+1)\]
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