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?

  • $\begingroup$ Where are your differentials? $\endgroup$ – Pedro Tamaroff Jun 29 '13 at 16:51
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    $\begingroup$ It doesn't appear that you actually took the antiderivative from step 5 to step 6 $\endgroup$ – David Mitra Jun 29 '13 at 16:52
  • $\begingroup$ 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. $\endgroup$ – Miguel Jun 29 '13 at 16:53
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    $\begingroup$ 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. $\endgroup$ – Lord_Farin Jun 29 '13 at 17:00

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

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

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



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


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 ;-)

  • $\begingroup$ thanks definititely sloppy in my handwriting. but you are absolutely right i forgot to take the integral. Long day thanks for the answer. $\endgroup$ – Miguel Jun 29 '13 at 17:09
  • $\begingroup$ You're welcome, Miguel! $\endgroup$ – Namaste Jun 29 '13 at 17:10
  • $\begingroup$ @amWhy: Looks good to me +1 $\endgroup$ – 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.

  • $\begingroup$ this seems like a clever solution unfortunately i dont understand it but i would definitely study this a little more. $\endgroup$ – Miguel Jun 29 '13 at 17:08

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


I'm going to show you a general method for integrals like this.

Consider the integral $$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$ $t=\sin^2x$:

$\therefore dt=2\sin x\cos x\ dx\\\therefore dx=\frac12t^{-1/2}(1-t)^{-1/2}dt\\\therefore x=0\mapsto t=0\\\therefore x=\pi/2\mapsto t=1$ $$I(a,b)=\int_0^1 t^{a/2}(1-t)^{b/2}\frac12t^{-1/2}(1-t)^{-1/2}dt$$ $$I(a,b)=\frac12\int_0^1 t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$ $$I(a,b)=\frac12\int_0^1 t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$ Note the definition of the Beta function: $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(b,a)$$ Thus $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}=I(b,a)$$ Where $\Gamma(x)$ is the Gamma Function.


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