Integrating a trigonometric integral with multiple variables

[utkarshakash]

Homework Statement

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

Homework Equations

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?

 

[Saitama]

Homework Statement

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

Homework Equations

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?

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.

And what about your previous thread? Did you get the answer?

 

[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.

 

[Saitama]

^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.

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.

 

[utkarshakash]

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.

And what about your previous thread? Did you get the answer?

Yes I got the answer of previous question.

 

[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.

 

[sankalpmittal]

Yes I got the answer of previous question.

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...

 

[Saitama]

This will eliminate x.

No. How do you conclude that?

 

[sankalpmittal]

No. How do you conclude that?

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

On adding,

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

Now take, a² cos² x+b² sin² x=t

 

[Saitama]

Then applying the property of definite integration,

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

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

 

[sankalpmittal]

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

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.

 

[Saitama]

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.

You ignored the denominator when you applied the property.

 

[utkarshakash]

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.

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.

 

[Saitama]

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.

Did you arrive at the answer?

 

[utkarshakash]

Did you arrive at the answer?

I got the answer by your method. Thanks!

 

[Saitama]

I got the answer by your method. Thanks!

Glad to help!

 

[sankalpmittal]

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.

Oh, I ignored the denominator when I applied the property !

However, when you integrate by parts, you can also divide both sides by cos⁴x, and then take tan²(x) = t. This will simplify the integrand to a mere identity.