Question Number 148137 by Willson last updated on 25/Jul/21

∫ ((3cos^2 (x)+1)/(sin^5 (x)))dx = ???

Answered by puissant last updated on 25/Jul/21

=∫((4cos^2 (x)+sin^2 (x))/(sin^5 (x)))dx=∫((4+tan^2 (x))/(tan^2 (x)sin^3 (x)))dx =∫(((4/(tan^2 (x)))+1)/(sin^3 (x)))dx=∫((4cos^2 (x)−3)/(sin^3 (x)))dx =∫((4cotan^2 (x)−(3/(sin^2 (x))))/(sin(x)))dx=∫((1/(sin^2 (x)))/(sin(x)))dx=∫(1/(sin^3 (x)))dx =∫(1/(sin(x)(1−cos^2 (x))))dx t=cos(x) ⇒ dt=−sin(x)dx⇒dx=−(dt/(sin(x))) ⇒I=−∫(dt/((1−t^2 )^2 )) =−∫((a/(1−t))+(b/((1−t)^2 ))+(c/(1+t))+(b/((1+t)^2 )))dt a=b=c=d=(1/4) ⇒I=(1/4)ln∣1−t∣−(1/(4(1−t)))−(1/4)ln∣1+t∣+(1/(4(1+t)))+C I=(1/4)ln∣((1−cos(x))/(1+cos(x)))∣−(1/(2sin^2 (x)))+C..