MHB Integrating $\sin^4 x$ to get $\cos^2 2x$

[karush]

$$\int \sin^4\left({x}\right) dx
\implies\int \left(\sin^2 \left({x}\right)\right)^2 dx$$

$$\implies \frac{1}{4}\int\left(1-\cos\left({2x}\right)\right)^2 dx

\implies \frac{1}{4}\int\left(1-2\cos\left({2x}\right)+\cos^2 \left({2x}\right)\right)dx $$

Got this far...hope ok

 

[Greg]

You've got too many $\dfrac14$'s in the first integral on the second line; everything else looks ok. Alternatively, you could rewrite the integrand as $\sin^2x(1-\cos^2x)$, which gives

$$\int\sin^2(x)\,dx-\dfrac14\int\sin^2(2x)\,dx$$

Then use the half-angle identities to simplify. That may be a little easier.By the way, why did you title this thread "abs limit"?

 

[Prove It]

$$\int \sin^4\left({x}\right) dx
\implies\int \left(\sin^2 \left({x}\right)\right)^2 dx$$

$$\implies \frac{1}{4}\int\left(1-\cos\left({2x}\right)\right)^2 dx

\implies \frac{1}{4}\int\left(1-2\cos\left({2x}\right)+\cos^2 \left({2x}\right)\right)dx $$

Got this far...hope ok

This is fine, now you are going to need to use a double angle formula again on the final term...

 

[karush]

Can this be done with the chain rule and not expansion

 

[Prove It]

Can this be done with the chain rule and not expansion

No, if you were going to use the chain rule you need a multiple of the derivative of cos(2x)...