5
$\begingroup$

I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.

$$\int\limits_0^{\frac{\pi}{2}}\sin^5(x)\cos^{14}(x)dx$$

My work:

$$\int\limits_0^{\frac{\pi}{2}}[\sin^2(x)]^2\cos^{14}(x)\sin(x)dx$$

$$\int\limits_0^{\frac{\pi}{2}}[1-\cos^2(x)]^2\cos^{14}(x)\sin(x)dx$$

substitute $\cos(x)$ for $u$.
$du=\sin(x)dx$

$$\int\limits_0^{\frac{\pi}{2}}(1-u^2)^2u^{14}du$$

$$\int\limits_0^{\frac{\pi}{2}}(1-2u+u^4)u^{14}du$$

$$\int\limits_0^{\frac{\pi}{2}}u^{14}-2u^{16}+u^{18}du$$

$$\frac{1}{15}u^{15}-\frac{2}{17}u^{17}+\frac{1}{19}u^{19}\Bigg|_{-1}^0=$$

$$\frac{1}{15}\cos^{15}\left(\frac{\pi}{2}\right)-\frac{2}{17}\cos^{17}\left(\frac{\pi}{2}\right)+\frac{1}{19}\cos^{19}\left(\frac{\pi}{2}\right) - \left(\frac{1}{15}\cos^{15}(0)-\frac{2}{17}\cos^{17}(0)+\frac{1}{19}\cos^{19}(0)\right)$$

$$0-0+0-(1-1+1)$$

$$-1+1-1=-1$$

The correct answer is $\frac{8}{4845}$

Where did I go wrong?

$\endgroup$
4
$\begingroup$

It is in fact $u=-\cos x$ and by substitution we get $$\frac{1}{15}u^{15}-\frac{2}{17}u^{17}+\frac{1}{19}u^{19}\Bigg|_{-1}^0={1\over 15}-{2\over 17}+{1\over 19}={8\over 4845}$$You where only wrong in substitution...

$\endgroup$
  • $\begingroup$ And I forgot to multiply the 1 by the fractions... I feel stupid. Thank you for the help. $\endgroup$ – Haruku Mar 18 at 11:09
  • $\begingroup$ Your welcome. Good luck!! $\endgroup$ – Mostafa Ayaz Mar 18 at 11:10
1
$\begingroup$

$$u = \cos x$$ $$du = -\sin x \,dx$$

(and not $du = \sin x \,dx$, as you wrote).

$\endgroup$
1
$\begingroup$

There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is $$u=\cos(x)$$ then the limit of integration change to

$$\int\limits_{\color{red}{0}}^\color{red}{{\frac{\pi}{2}}}f(\cos(x))(-\sin(x))dx=\int\limits_{\color{red}{\cos0}}^\color{red}{{\cos\frac{\pi}{2}}}f(u)du={F(u){\LARGE|}}_{\color{red}{\cos0}}^\color{red}{{\cos\frac{\pi}{2}}}=F(\cos(x)){\LARGE|}_{\color{red}{0}}^\color{red}{{\frac{\pi}{2}}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.