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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.

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    $\begingroup$ Does nothing jump out to you about $\sqrt{u^{2}}$? $\endgroup$ – preferred_anon Mar 6 '18 at 20:52
  • $\begingroup$ 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. $\endgroup$ – gbgult Mar 6 '18 at 20:59
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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: $\int_{1}^{\sqrt{2}} u^2 du = [\frac{u^3}{3}]_{1}^{\sqrt{2}} = \frac{2^{\frac{3}{2}} - 1}{3}$.

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  • $\begingroup$ Thanks for the answer! One question though; shouldn't du be equal to: $-\frac{sin(x)}{cos^2(x)} dx$ ? $\endgroup$ – gbgult Mar 6 '18 at 21:16
  • $\begingroup$ Oh, nevermind, forgot the reciprocal rule. $\endgroup$ – gbgult Mar 6 '18 at 21:22
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your upper limit in $$\int_1^\frac{\sqrt2}{2}{u\sqrt{u^2} du}$$ should have been $\sqrt 2$.

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

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

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

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