Answer to Question #233072 in Calculus for moe

Question

Evaluate the integral $\displaystyle{\int_{0}^{\pi/2}\sin{(x)}\ln{\sec{(x)}}dx}$

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4. 8

Answer 1

$\displaystyle{\int_{0}^{\pi/2}\sin{(x)}\ln{\sec{(x)}}dx}$

Solution

$\displaystyle{\int\sin{(x)}\ln{\sec{(x)}}dx}$

Integrate by parts $\displaystyle\int fg'=fg-\int f'g$

$f=ln(sec(x))$

$f'=tan(x)$

$g'=sin(x)$

$g=-cos(x)$

Hence from $\displaystyle\int fg'=fg-\int f'g$

$=-cos(x)ln(sec(x))-\displaystyle\int-cos(x)tan(x)dx$

Now solving ;

$\displaystyle\int-cos(x)tan(x)dx$

Apply linearity;

$=-\displaystyle\int cos(x)tan(x)dx$

Now solving;

$\displaystyle\int cos(x)tan(x)dx$

Rewrite/ Simplify using trigonometric/hyperbolic identities

$=\displaystyle\int sin(x)dx$

This is the standard integrals:

$=-cos (x)$

Plug in solved integrals;

$=-cos(x)ln(sec(x))-\displaystyle\int-cos(x)tan(x)dx$

$=(-cos(x)ln(sec(x))-cos(x)$

The problem is solved

$\displaystyle{\int\sin{(x)}\ln{\sec{(x)}}dx}$

$=[-cos(x)ln(sec(x))-cos(x)+C]$

Simplify

$=cos(x)(ln(|cos(x)|)-1)+C$

Hence

 $\displaystyle{\int_{0}^{\pi/2}\sin{(x)}\ln{\sec{(x)}}dx}= [cos(x)(ln(|cos(x)|)-1)+C]_0^{\frac{π}{2}}$

$=[cos(\frac{π}{2})(ln(|cos(\frac{π}{2})|)-1)+C]- [cos(0)(ln(|cos(0)|)-1)+C]= 1$