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  • Stuck here, help me understand: - Integral Calculus - JEE Main-8

\int_{0}^{\pi/2}\frac{sinx}{1+cosx+sinx}dx

  • Option 1)

    \frac{\pi}{4}

  • Option 2)

    \frac{\pi}{4}+log\sqrt{2}

  • Option 3)

    \frac{\pi}{4}-log\sqrt{2}

  • Option 4)

    \frac{\pi}{4}-log{2}

 

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

Integrals in the form of     \int \frac{p\cos x+q\sin x }{a\cos x+b\sin x}dx

- wherein

Working rule :

p\cos x+q\sin x=A\cdot \frac{\mathrm{d} }{\mathrm{d} x}(a\cos x+b\sin x)+B(a\cos x+b\sin x) 

Find A and B by comparing sinx and cosx 

 

 I = \int\limits_0^{\pi /2} {\frac{{\sin \frac{x}{2}}}{{\sin \frac{x}{2} + \cos \frac{x}{2}}}dx = 2\int\limits_0^{\pi /4} {\frac{{\sin x}}{{\sin x + \cos x}}dx} }

Let\: \[\sin x = A(\sin x + \cos x) + B(\cos x - \sin x)\]

A - B = 1\: \: and\: \: A + B = 0,A = \frac{1}{2},B = - \frac{1}{2}

I = 2 \times \frac{1}{2} \times \frac{\pi }{4} - 2 \times \frac{1}{2}\left[ {\log (\sin x + \cos x)} \right]_0^{\frac{\pi }{4}}=\frac{\pi }{4} - \log \sqrt 2


Option 1)

\frac{\pi}{4}

Option 2)

\frac{\pi}{4}+log\sqrt{2}

Option 3)

\frac{\pi}{4}-log\sqrt{2}

Option 4)

\frac{\pi}{4}-log{2}

Posted by

prateek

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