"I am back" says Integration - Part 2

Calculus Level 5

$\large \int \frac{\sqrt{\cos(2x)}}{\sin(x)} \, dx$

If the integral above can be expressed as $\sqrt{a}\log(\sqrt{b}\cos{x} + \sqrt{\cos(cx)}) - \log(d\cot{x} + ((\cot{x})^e - g)^f)$ , for constants $a,b,c,d,e,g$ and $f$ , evaluate $\frac{abc}{dg} + ef + 11$ .

Note: $f$ is not an integer. And all these constants may not be distinct.

The answer is 20.

2 solutions

Rishabh Singhal
Aug 19, 2015

Let $I=\int { \frac { \sqrt { cos(2x) } }{ sinx } dx\quad } =\int { \frac { cos(2x) }{ sinx\sqrt { cos(2x) } } } dx =\int { \frac { 1-2\sin ^{ 2 }{ x } }{ sinx\sqrt { cos(2x) } } } dx =\int { \frac { 1 }{ sinx\sqrt { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } } } dx\quad -2\int { \frac { sinx }{ \sqrt { 2\cos ^{ 2 }{ x } -1 } } dx } } =\int { \frac { \csc ^{ 2 }{ x } }{ \sqrt { \cot ^{ 2 }{ x\quad -1 } } } dx\quad -\frac { 2 }{ \sqrt { 2 } } } \int { \frac { sinx }{ \sqrt { \cos ^{ 2 }{ x } -1 } } dx } =-\int { \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } -\sqrt { 2 } \int { \frac { -ds }{ \sqrt { { s }^{ 2 }-(\frac { 1 }{ \sqrt { 2 } } )^{ 2 } } } } \quad where\quad t=cotx\quad and\quad s=cosx ; =-\ln { \left| t+\sqrt { t^{ 2 }-1 } \right| } +\sqrt { 2 } \ln { \left| \sqrt { 2 } s\quad +\sqrt { 2s^{ 2 }-1 } \right| } ] = \sqrt{2}\log(\sqrt{2}\cos{x} + \sqrt{\cos(2x)}) - \log(\cot{x} + \sqrt{\cot^2{x} - 1})$

Rajdeep Dhingra
Feb 24, 2015

The answer is $I = \sqrt{2}\log(\sqrt{2}\cos{x} + \sqrt{\cos(2x)}) - \log(\cot{x} + \sqrt{\cot^2{x} - 1})$

Moderator note:

Can you provide a proper solution?

We would be much more interested to know how you solved it..

Ayush Garg - 5 years, 11 months ago

"I am back" says Integration - Part 2

The answer is 20.

2 solutions

Moderator note:

0 pending reports