Integral u-substitution

Solve this integral: $$\int_{0}^{\pi/4}\frac{\sin\left(x\right)}{\cos^{3}\left(x\right)} \,\sqrt{\,\cos^{-2}\left(\, x\, \right)\,}\,\,\mathrm{d}x$$

Use $u$-substitution where $u = \dfrac {1}{\cos(x)}$.

I get to the point where my integral is: $$\int_1^\frac{\sqrt2}{2}{u\sqrt{u^2} du}$$ but don't really know where to go from there. I might have done something wrong on the way as well, not $100$% sure about that.

• Does nothing jump out to you about $\sqrt{u^{2}}$? – preferred_anon Mar 6 '18 at 20:52
• Oh, that's stupid of me, haha. The problem is that it's the wrong answer then so I've definitely done something wrong on the way. – gbgult Mar 6 '18 at 20:59

The upper limit of the integral in $u$ should be $\sqrt{2}$ because when $x = \frac{\pi}{4}$, $u = \frac{1}{\cos x} = \sqrt{2}$. Apart from that, you are on the right track. Here is my solution:

$\int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos^3 x} \frac{1}{\sqrt{\cos^2 x}} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos^3 x} \frac{1}{\cos x} dx$.

After the substitution $u = \frac{1}{\cos x}$, $du = \frac{\sin x}{\cos^2 x}dx$, we have: $\int_{1}^{\sqrt{2}} u^2 du = [\frac{u^3}{3}]_{1}^{\sqrt{2}} = \frac{2^{\frac{3}{2}} - 1}{3}$.

• Thanks for the answer! One question though; shouldn't du be equal to: $-\frac{sin(x)}{cos^2(x)} dx$ ? – gbgult Mar 6 '18 at 21:16
• Oh, nevermind, forgot the reciprocal rule. – gbgult Mar 6 '18 at 21:22

your upper limit in $$\int_1^\frac{\sqrt2}{2}{u\sqrt{u^2} du}$$ should have been $\sqrt 2$.

Thus we have $$\int_{0}^{\pi/4}\frac{\sin\left(x\right)}{\cos^{3}\left(x\right)} \,\sqrt{\,\cos^{-2}\left(\, x\, \right)\,}\,\,\mathrm{d}x=$$

$$\int_1^{\sqrt2}{u\sqrt{u^2} du}=$$

$$\int_1^{\sqrt2}u^2 du =\frac{2\sqrt 2 - 1}{3}$$