Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

im asked to find the limited integral here but unfortunately im floundering can someone please point me in the right direction? $$\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $$

step 1 brake up sin and cos so that i can use substitution $$\int_0^\frac{\pi}{2} \sin^7(x) \cos^4(x) \cos(x) \, dx$$ step 2 apply trig identity $$\int_0^\frac{\pi}{2} \sin^7x\ (1-\sin^2 x)^2 \, dx$$
step 3 use $u$-substitution $$ \text{let}\,\,\, u= \sin(x)\ du=\cos(x) $$ step 4 apply use substitution $$\int_0^\frac{\pi}{2} u^7 (1-u^2)^2 du $$ step 5 expand and distribute and change limits of integration $$\int_0^1 u^7-2u^9+u^{11}\ du $$ step 6 integrate $$(1^7-2(1)^9+1^{11})-0$$ i would just end up with $1$ however the book answer is $$\frac {1}{120}$$

how can i be so far off?

share|cite|improve this question
Where are your differentials? – Pedro Tamaroff Jun 29 '13 at 16:51
It doesn't appear that you actually took the antiderivative from step 5 to step 6 – David Mitra Jun 29 '13 at 16:52
OMG i feel so stupid right now. Long day guys ive been looking at this for too long. My apologies. thanks for pointing out my dumb mistake. – Miguel Jun 29 '13 at 16:53
Style protip: Fractions $\frac ab$ in integral bounds can become difficult to read; consider using $a/b$ instead. See this thread for more (Math.SE-specific) TeX tips. – Lord_Farin Jun 29 '13 at 17:00
up vote 4 down vote accepted

$$\int_0^\frac\pi2\sin^7x\cos^5xdx=\int_0^\frac\pi2\sin^7x\cos^4x\cos xdx=\int_0^\frac\pi2\sin^7x(1-\sin^2x)^2\cos xdx$$

$$=\int_0^1 u^7(1-u^2)^2 du (\text{ Putting }\sin x=u)$$

$$=\int_0^1 (u^7-2u^9+u^{11}) du$$



$$=\frac{15-24+10}{120}=\frac1{120} $$

share|cite|improve this answer

You forgot to integrate between step $(5)$ and $(6)$! $$\int_0^1 \left(u^7-2u^9+u^{11}\right)\ du \quad =\quad \left(\frac{u^8}{8} -\frac{u^{10}}{5} + \frac{u^{12}}{12}\right)\Big|_0^1 = \frac 18 - \frac 15 +\frac 1{12} = \frac 1{120}$$

You're work was fine, otherwise (you left out $\,dx$ from your earlier integrals, and the factor $\cos x$, which turns out to be $\,du$ and so accommodated in the substitution in the second step), but I think your primary lapse was simply forgetting to integrate before evaluating ;-)

share|cite|improve this answer
thanks definititely sloppy in my handwriting. but you are absolutely right i forgot to take the integral. Long day thanks for the answer. – Miguel Jun 29 '13 at 17:09
You're welcome, Miguel! – amWhy Jun 29 '13 at 17:10
@amWhy: Looks good to me +1 – Amzoti Jun 30 '13 at 0:29

Another approach using $$\frac{m! n!}{(m+n+1)!}=\operatorname{B}(m+1,n+1)=2\int_0^{\frac{\pi}{2}}\cos^{2m+1}x\sin^{2n+1}x \, dx.$$

In our case $$\int_0^\frac{\pi}{2} \sin^7x \cos^5x \, dx=\frac{\operatorname{B}(3,4)}{2}=\frac{1}{2}\frac{2! 3!}{6!}=\frac{1}{120}.$$

Here $\operatorname{B}$ denotes Beta function.

share|cite|improve this answer
this seems like a clever solution unfortunately i dont understand it but i would definitely study this a little more. – Miguel Jun 29 '13 at 17:08

step 6 is wrong, check it again, it should be $$\frac{u^8}{8} + \frac{u^{12}}{12} - \frac{u^{10} }{5}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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