Efficient Methods for Evaluating Trig Integrals: Using Half Angle Formula

[courtrigrad]

1. \int^{\frac{\pi}{4}}_{0} sin^{4}x\ cos^{2} x \ dx. Would it work to use the half angle formula for both terms? I did this, it took very long. Any quick methods to evaluate this?

Thanks

 

[quasar987]

After 5 quick sketchy integrations by parts on a scrap of paper, I get an answer. But I don't guaranty it's not a ghost. You can try if you want. The first step is taking u=sin^4(x)cos(x) dv=cos(x)dx.

At the end, I get the integral of sin²(x) which is 0.5(x-cosxsinx)

 

[VietDao29]

Yes, you can use Power-reduction formulae to evaluate this integral. It'll be a little bit messy, however.
Or you can do it this way. Here's my approach:
Let
A = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx
And we define another definite integral:
B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx
Now, if we sum the 2 integrals above, we'll have:
A + B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx + \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x (\sin ^ 2 x + \cos ^ 2 x) dx
= \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x dx = \frac{1}{4} \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 (2x) dx = ...
And we subtract B from A to get:
A - B = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 4 x \cos ^ 2 x dx - \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 4 x dx = \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x (\sin ^ 2 x - \cos ^ 2 x) dx
= - \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 x \cos ^ 2 x \cos (2x) dx = - \frac{1}{4} \mathop{\int} \limits_{0} ^ {\frac{\pi}{4}} \sin ^ 2 (2x) \cos (2x) dx = ...
The two integrals above are easier to evaluate than the original one, right?
Having A + B, and A - B, can you work out A? :)

 

[quasar987]

That was ingenious!