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

  • Option 1)

    \frac{\Pi }{4}

  • Option 2)

    \frac{\Pi }{4}+log\sqrt2

  • Option 3)

    \frac{\Pi }{4}-log\sqrt2

  • Option 4)

    \frac{\Pi }{4}-log2

 

Answers (1)

As we learnt

Properties of Definite Integration -

\int_{0}^{2a}f(x)dx= \int_{0}^{a}\left [ f(x)+f(-x) \right ]dx

= \left\{\begin{matrix} 2\int_{0}^{a}f(x)dx & if f(2a-x)=f(x)\\ 0 &if f(2a-x)=-f(x) \end{matrix}\right.

-

 

 I=\int_{0}^{\pi/2}\frac{sin\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}}dx=2\int_{0}^{\pi/4}\frac{sinx}{sinx+cosx}dx

 Let \: \: sinx=A(sinx+cosx)+B(cosx-sinx)

A-B=1\: \: and\: \: A+B=0;A=1/2;B=-1/2

I=2*\frac{1}{2}*\frac{\pi }{4}-2*\frac{1}{2}\left [log(sinx+cosx) \right ]_{0}^{\pi /4}=\frac{\pi }{4}-log\sqrt{2}


Option 1)

\frac{\Pi }{4}

Option 2)

\frac{\Pi }{4}+log\sqrt2

Option 3)

\frac{\Pi }{4}-log\sqrt2

Option 4)

\frac{\Pi }{4}-log2

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subam

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