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I have to calculate this Integral -

$$\int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}dx$$

I have no idea how to start, Hint someone ? Thanks.

gt6989b
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NM2
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    This one occurs every now and then. If you do $u=\pi/2-x$ you get the same integral but with $\cos$ instead of $\sin$ in the numerator. Add them and you get the integral of the function $1$. This should be a duplicate, but the $7/2$ can be changed to more or less whatever, so it is not so easy to find the other ones... – mickep Jan 19 '16 at 19:49
  • How will you get the integral of $1$ ? – NM2 Jan 19 '16 at 20:07
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    @Noam $$\frac{s}{s+c} + \frac{c}{s+c} = 1$$ – gt6989b Jan 19 '16 at 20:13
  • http://math.stackexchange.com/questions/605673/integrate-int-0-pi-2-frac11-tan-alphax-mathrmdx/605713#605713 – Ron Gordon Jan 19 '16 at 20:13

1 Answers1

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Use the following: $$\int_a^{b} f(x) \, dx = \int_a^{b} f(a+b-x) \, dx$$ This property can be proven easily by substituting $a+b-x=t$.

Now consider $f(x)=\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}$ and $a=0$, $b=\pi/2$, to get $$I=\int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}dx=\int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}\left(\frac{\pi}{2}-x\right)}{\sin^{7/2}\left(\frac{\pi}{2}-x\right)+\cos^{7/2}\left(\frac{\pi}{2}-x\right)}dx.$$ Thus we have \begin{align*} I & = \int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}dx\\ \text{and also }\\ I & = \int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}\left(\frac{\pi}{2}-x\right)}{\sin^{7/2}\left(\frac{\pi}{2}-x\right)+\cos^{7/2}\left(\frac{\pi}{2}-x\right)}dx= \int _0^{\frac{\pi }{2}}\:\frac{\cos^{7/2}x}{\cos^{7/2}x+\sin^{7/2}x}dx. \end{align*} Adding the two to get $$2I=\int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}x+\cos^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}dx=\int_0^{\frac{\pi}{2}}1 \, dx=\frac{\pi}{2}.$$ Thus $I=\frac{\pi}{4}$.

NOTE: as indicated in one of the comments, $7/2$ has no significance, it could have been replaced by another suitable exponent and still the answer would be the same.

Anurag A
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  • Ok,Thanks. But, Why $\int_a^{b} f(x) , dx = \int_a^{b} f(a+b-x) , dx$ ? – NM2 Jan 19 '16 at 20:34
  • @Noam if you read my solution, I have also given the hint of proving it by using the substitution $a+b-x=t$. – Anurag A Jan 19 '16 at 20:36
  • @Noam using $a+b-x=t$, you can have $\frac{dx}{dt}=-1$ and for the limits when $x=a$, then $t=b$, when $x=b$, then $t=a$. With this $\int_a^bf(a+b-x) , dx=-\int_b^a f(t) dt=\int_a^b f(t) , dt.$ – Anurag A Jan 19 '16 at 20:39
  • If i use substitution, I will get. $\int _0^{\frac{\pi :}{2}}:\frac{\sin ^{7/2}x}{\sin ^{7/2}x+\cos ^{7/2}x}dx:=:-:\int _0^{\frac{\pi ::}{2}}:\frac{cos:^{7/2}t}{cos:^{7/2}t+sin:^{7/2}t}dt$ – NM2 Jan 19 '16 at 20:42
  • @Noam there will be no negative sign because the limits will be reversed. – Anurag A Jan 19 '16 at 20:43
  • Ok,So - $\int _0^{\frac{\pi :}{2}}:\frac{\sin ^{7/2}x}{\sin ^{7/2}x+\cos ^{7/2}x}dx:=::\int _{\frac{\pi }{2}}^0:\frac{cos:^{7/2}t}{cos:^{7/2}t+sin:^{7/2}t}dt$

    But then(If the limits are reversed), How can i adding them together ?

     – NM2 Jan 19 '16 at 20:44
  • Let us continue this discussion in chat. – NM2 Jan 19 '16 at 20:45