Question

Ex. (7) solve the following definite integration​

Answer 1

Answer:

(7) Integral 0 to π x tan x/ sec x cosec x

= Integral 0 to π x (sinx/cosx * sec x * cosec x )

= Integral 0 to π x( sin x/cosec x)

= Integral 0 to π x sin^2 x dx

Let,

I = Integral 0 to π x sin^2 x dx

Now by using the property that

integral a to b f(x) dx = Integral a to b f(a+b-x) dx

So,

I = integral 0 to π x sin^2 x dx

which by property is also equal to

I = Integral 0 to π (π-x) sin^2(π-x) dx

= Integral 0 to π. πsin^2 x dx - Integral 0 to π x sin^2 x dx

Because, sin(π-x) = sin x

But ,on RHS we recognise - Integral 0 to π x sin^2 x dx as - I

so, equation becomes,

I = Integral 0 to π (πsin^2 x dx ) - I

or,

I = 1/2(π) Integral 0 to π sin^2 x dx

Now, we have to evaluate the integral 0 to π sin^2 x dx

So, sin^2 x = 1/2(1-cos 2x)

So,

Integral 0 to π sin^2 x dx = 1/2* Integral 0 to π (1-cos 2x) dx

= 1/2(x - sin2x/2) limits from 0 to π

= 1/2(π -0) = π/2

So, finally

I = π/2(Integral 0 to π sin^2 x dx = (π/2)(π/2) = π^2/4

hope this helps you !

Answer 2

Answer : $\dfrac{π^2}{4}$

$\bold{I\;hope\;it\;will\;be\;helpful!}$