Integration By Parts Is Crazy

Use this property of integrals. $\int_0^kf(x)\text{ }dx=\int_0^kf(k-x)\text{ }dx$ Therefore, we can show the integrals like this. $\int_0^\frac{\pi}{2}\sin^{2014}x-\cos^{2014}x\text{ }dx=\int_0^\frac{\pi}{2}\sin^{2014}\left(\dfrac{\pi}{2}-x\right)-\cos^{2014}\left(\dfrac{\pi}{2}-x\right)\text{ }dx$ $\int_0^\frac{\pi}{2}\sin^{2014}x-\cos^{2014}x\text{ }dx=\int_0^\frac{\pi}{2}\cos^{2014}x-\sin^{2014}x\text{ }dx$ $\int_0^\frac{\pi}{2}\sin^{2014}x-\cos^{2014}x\text{ }dx=-\int_0^\frac{\pi}{2}\sin^{2014}x-\cos^{2014}x\text{ }dx$ The integral is equal to its additive inverse. The only number that has this property is $\boxed{0}$

$\int_{0}^{\frac{\pi}{2}} (\sin^{2014}x - \cos^{2014}x)dx = \int_{0}^{\frac{\pi}{2}} (\sin^{2014}x)dx - \int_{0}^{\frac{\pi}{2}} (\cos^{2014}x)dx$

In the first integral, use the following variable substitutions:

$x=\frac{\pi}{2}-u$

$dx= -du$

$x=\frac{\pi}{2} \Rightarrow u=0$

$x=0 \Rightarrow u=\frac{\pi}{2}$

This changes things as follows:

$\int_{\frac{\pi}{2}}^{0} (\sin^{2014}(\frac{\pi}{2}-u))(-du) - \int_{0}^{\frac{\pi}{2}} (\cos^{2014}x)dx$

$=\int_{0}^{\frac{\pi}{2}} (\sin^{2014}(\frac{\pi}{2}-u))du - \int_{0}^{\frac{\pi}{2}} (\cos^{2014}x)dx$

$=\int_{0}^{\frac{\pi}{2}} (\cos^{2014}u)du - \int_{0}^{\frac{\pi}{2}} (\cos^{2014}x)dx$

$=0 \quad \Box$

We can obtain the graph of $\cos^{2014}x$ by reflecting the graph of $\sin^{2014}x$ about the vertical line $x=\frac{\pi}{4}$ , since $\cos^{2014}x=\sin^{2014}(x-\frac{\pi}{2})$ . Thus the two integrals (the "areas under the curve") are the same, and the answer is 0.

Set u=pi/2-x==> int (cos(u)^2014-sin(u)^2014, u, 0, pi/2)=int(sin(x)^2014-cos(x)^2014,x,0, pi/2)==> the answer to the question is half of the sum of above integrals which is zero

just integrate the function to get: ((sinx)^2015)/(2015cosx)-((cosx)^2015)/(2015sinx). then evalute at the limits to give [1/0-0/2015]-[0/2015-1/0]. normally this would be a bit of a problem with this being infinite-infinite. but you can see that they're they're the same kinds of infinities. since they come from functions with the same powers/coeficients whatever. so they cancel to give an answer of 0.

the given limits are 0 and pie so that if integrate sinx-cosx with their powers as 2014 we get sin^2015 x/2015 cosx-cos^2015x/2015 (-sinx) now we apply limits then sin 0 is 0 and coso is 1 sin 90 is 1and cos90 is 0 if we substitute them we get zero

(sin x)^2014 - (cos X)^2014

write this in the form of a^2-b^2

which will be {(sinx)^1007}^2 - {(cosx)^1007}^2, where a= (sinx)^1007, and b= (cosx)^1007

we know, (sinx)^2-(cosx)^2 = cos2x,

substituting sinx from the previous equation with (sinx)^1007, and cosx from the previous equation with (cosx)^1007

we get {(sinx)^1007}^2 - {(cosx)^1007}^2= (cos 2x)^1007

Integrating (cos 2x)^1007 will give [{(sin2x)}^1007]/ (1007*2)

Then just put in the intervals, and find out the answer!! :)

The area bounded by both sin (x) and cos (x) from 0 to (pi/2) is same. Therefore when subtracted the integral becomes zero.