0
$\begingroup$

Question

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2x\cdot \log \sin x dx$$

I put $\sin x = t $

Then $\cos x = \sqrt{1-t^2}$

Also $\cos x dx = dt$.

But I am getting very large and wrong answer. If you want more information please let me know.

Can someone help me with this one.

1 Answers 1

1

Using integration by parts:

$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos\left(2x\right)\ln\left(\sin\left(x\right)\right)\space\text{d}x=\left[\frac{\sin\left(2x\right)\ln\left(\sin\left(x\right)\right)}{2}\right]_\frac{\pi}{4}^\frac{\pi}{2}-\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{\cos\left(x\right)\sin\left(2x\right)}{2\sin\left(x\right)}\space\text{d}x\tag1$$

Now, see that:

$$\left[\frac{\sin\left(2x\right)\ln\left(\sin\left(x\right)\right)}{2}\right]_\frac{\pi}{4}^\frac{\pi}{2}=\frac{\sin\left(2\cdot\frac{\pi}{2}\right)\ln\left(\sin\left(\frac{\pi}{2}\right)\right)}{2}-\frac{\sin\left(2\cdot\frac{\pi}{4}\right)\ln\left(\sin\left(\frac{\pi}{4}\right)\right)}{2}=\frac{\ln2}{4}\tag2$$

And for the integral, substitute $\text{u}=\tan\left(x\right)$:

$$\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{\cos\left(x\right)\sin\left(2x\right)}{2\sin\left(x\right)}\space\text{d}x=\frac{1}{2}\int_1^\infty\frac{1}{\left(1+\text{u}^2\right)^2}\space\text{d}\text{u}=\frac{1}{2}\int_1^\infty\frac{1}{1+\text{u}^2}\space\text{d}\text{u}-\frac{1}{4}\tag3$$

  • 2
    The second integrand simpifies to $\cos^2x$. – 2017-02-05
  • 0
    ...Which is also $\frac12\cos(2x)+\frac12$, whose primitives are direct. – 2017-02-06