# Evaluate this integral

1. Jun 23, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
$\displaystyle \int^{\pi /2}_0 \dfrac{xcosxsinx}{(a^2 cos^2 x+b^2 sin^2 x)^2} dx$

2. Relevant equations

3. The attempt at a solution

2I = $\displaystyle \int^{\pi /2}_0 \dfrac{xcosxsinx}{(a^2 cos^2 x+b^2 sin^2 x)^2} + \dfrac{(\pi /2 -x)cosxsinx}{(a^2 sin^2 x+b^2 cos^2 x)^2} dx$

How to simplify after this?

2. Jun 23, 2013

### Saitama

I couldn't think of a smarter way but the following should work.

You can write the given integral as
$$\int_0^{\pi/2} \frac{x\tan x \sec^2x}{(a^2+b^2\tan^2x)^2}dx$$
Use the substitution $\tan x=t$ and then integrate by parts.

3. Jun 23, 2013

### lurflurf

^Are you trying to get the antiderivative? It is easy to get rid of the x in front by integrating by parts, but the new integral is hard. Perhaps the bounds can be used, but the a and b break the symmetry.

4. Jun 23, 2013

### Saitama

Yes, I am trying to find the antiderivative. The new integral isn't hard but only a bit lengthy (I think).

The new integral after substitution,
$$\int_0^{\infty} \frac{\arctan t\cdot t}{(a^2+b^2t^2)^2}dt$$

This can be solved using integration by parts.

5. Jun 24, 2013

### utkarshakash

Yes I got the answer of previous question.

6. Jun 24, 2013

### Saitama

I found a shorter method.

Solve the given integral using integration by parts with $u=x$ and $v$ be the rest i.e
$$x\int \frac{\sin x \cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int \left(\int \frac{\sin x \cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right)dx$$

Use the substitution, $(a^2\cos^2x+b^2\sin^2x)=t$ that should give the answer.

7. Jun 24, 2013

### sankalpmittal

This question is very easy by utkarshakash's method but he got afraid to see the big size of his integrand. :p

I too at first thought would use his method.

After simplifying you get,

2I=....

This will eliminate x. No need to integrate by parts. :)

Hint: Then use the substitution as Pranav suggested. Take sqrt(denominator)=t...

Last edited: Jun 24, 2013
8. Jun 24, 2013

### Saitama

No. How do you conclude that?

9. Jun 25, 2013

### sankalpmittal

Isn't is obvious ?

Let

I=π/20∫{xcosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

Then applying the property of definite integration,

I=π/20∫{(π/2-x)cosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

2I= I=π/20∫{π/2cosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

Now take, a2 cos2 x+b2 sin2 x=t

10. Jun 25, 2013

### Saitama

$$\cos(\pi/2-x)=\sin x$$

Last edited: Jun 25, 2013
11. Jun 25, 2013

### sankalpmittal

So what ?

cos x changes to sin x but sin x also changes to cos x. The net result remains, what I typed in my previous post. I have done such type of questions.

12. Jun 25, 2013

### Saitama

You ignored the denominator when you applied the property.

13. Jun 25, 2013

### utkarshakash

No. You are going the wrong way. Note that though cos x and sin x interchange, the coefficents are also reversed. $a^2 cos^2 x + b^2 sin^2 x$ changes to $a^2 sin^2 x + b^2 cos^2 x$. Now you can't just add the numerator simply. You know what I mean.

14. Jun 25, 2013

### Saitama

Did you arrive at the answer?

15. Jun 25, 2013

### utkarshakash

16. Jun 25, 2013