NIMCET 2009 Question Paper with solutions

© examsiri.com

Question : 6 of 120

Marks: +1, -0

The value of

∫↙{0}↖{π} \;{x \sin x}/{1+\cos ^{2} x} d x

is

$\;{π^{2}}/{3}$
$\;{π^{2}}/{4}$
$\;{π^{2}}/{6}$
$\;{π^{2}}/{2}$

Solution:

I=∫↙{0}↖{π} \;{x \sin x}/{1+\cos ^{2} x} d x=∫↙{0}↖{π} \;{(π-x) \sin x}/{1+\cos ^{2} x} d x

2 I=π ∫↙{0}↖{π} \;{\sin x}/{1+\cos ^{2} x} d x

∫↙{0}↖{2 a} f(x) d x=2 ∫↙{0}↖{a} f(x) d x \;\; \;\; \;\; ∵ ∵ f(2 ∆-x)=d(x)]

=2 π ∫↙{0}↖{π 2} \;{\sin x}/{1+\cos ^{2} x} d x

even function)

⇒ \;\; I=π ∫↙{0}↖{π \/ 2} \;{\sin x}/{1+\cos ^{2} x} d x

Let

\cos x=z \;\; ⇒-\sin x d x=d z

∴ \;\; I=π ∫↙{1}↖{0}-\;{d z}/{1+z^{2}}=π ∫↙{0}↖{1} \;{d z}/{1+z^{2}}

=π[\tan ^{-1} z]_{0}^{1}=π(\;{π}/{4})=\;{π^{2}}/{4}

© examsiri.com

Go to Question:

Prev Question

Next Question