Integrate (cos^3 x +cos ^5 x)/(sin^2 x+sin^4 x)?

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Answer 1 · 2018-06-29T06:27:18

$sin x - 2 csc x - 6 arctan sin x + C$ $sinx-2cscx-6arctan sinx+C$ .

Let, $I = \int \frac{{cos}^{3} x + {cos}^{5} x}{{sin}^{2} x + {sin}^{4} x} d x$ $I=int(cos^3x+cos^5x)/(sin^2x+sin^4x)dx$ ,

$= \int \frac{{cos}^{3} x (1 + {cos}^{2} x)}{{sin}^{2} x (1 + {sin}^{2} x)} d x$ $=int{cos^3x(1+cos^2x)}/{sin^2x(1+sin^2x)}dx$ ,

$= \int \frac{{cos}^{2} x (1 + {cos}^{2} x)}{{sin}^{2} x (1 + {sin}^{2} x)} cos x d x$ $=int{cos^2x(1+cos^2x)}/{sin^2x(1+sin^2x)}cosxdx$ ,

$=int{(1-sin^2x)(1+ul(1-sin^2x))}/{sin^2x(1+sin^2x)}cosxdx$ .

This shows that the substn. $sinx=t," so that, "cosxdx=dt,$

should work.

Hence, $I=int{(1-t^2)(2-t^2)}/{t^2(1+t^2)}dt=int(t^4-3t^2+2)/(t^4+t^2)dt$ ,

$=int{(t^4+t^2)-4t^2+2}/(t^4+t^2)dt$ ,

$=int{(t^4+t^2)/(t^4+t^2)-{2(2t^2-1)}/(t^4+t^2)}dt$ ,

$=int1dt-2int(2t^2-1)/{t^2(t^2+1)}dt$ ,

$=t-2int{(2t^2)/{t^2(t^2+1)}-1/{t^2(t^2+1)}}dt$ ,

$=t-2int{2/(t^2+1)-{(t^2+1)-t^2}/{t^2(t^2+1)}}dt$ ,

$=t-4int{1/(t^2+1)dt+2int{(t^2+1)/{t^2(t^2+1)}-t^2/{t^2(t^2+1)}}dt$ ,

$=2-4arctant+2int{1/t^2-1/(t^2+1)}dt$ ,

$=t-4arctant+2{-1/t-arctant}$ ,

$=t-2/t-6arctant$ .

Since, $t=sinx$ , we have,

$I=sinx-2cscx-6arctan sinx+C$ , as desired!

Enjoy Maths.!