RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer The value of\[\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}}dx,a>0\]is

    A)  \[a\pi \]

    B)  \[\frac{\pi }{2}\]

    C)  \[\frac{\pi }{a}\]

    D)  \[2\pi \]

    Correct Answer: B

    Solution :

     Let\[I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}}dx,a>0\]           ...(i) \[I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}}dx\] (put\[x=-x\])     ...(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{-\pi }^{\pi }{\frac{(1+{{a}^{x}})\cos x}{(1+{{a}^{x}})}}dx\] \[\Rightarrow \] \[2I=\int_{-\pi }^{\pi }{{{\cos }^{2}}x}dx\] \[\Rightarrow \] \[2I=\int_{-\pi }^{\pi }{\left( \frac{\cos 2x+1}{2} \right)}dx\] \[\Rightarrow \] \[2I=\frac{1}{2}\left[ \frac{\sin 2x}{2}+x \right]_{-\pi }^{\pi }\] \[\Rightarrow \] \[2I=\frac{1}{2}(\pi +\pi )\] \[\Rightarrow \] \[I=\frac{\pi }{2}\]

adversite



LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos