Question

Find:

`intsec^3 thetaxxd theta`

Answer 1

Let `I = intsec^3 thetaxxd theta`

Integrating by parts, we have u = sec θ, v = sec²θ

So, `I=intu.v.d theta`

`I = sec theta int sec^2theta.d theta-int {{d(sec theta))/(d theta) . int sec^2 theta.d theta} d theta`

`I = sec theta .tan theta - int sec theta. tan theta. tan theta. d theta`

`I = sec theta.tan theta - int sec theta.tan^2theta. d theta`

`I = sec theta. tan theta - int sec theta.(sec^2 theta - 1). d theta`

`I = sec theta.tan theta - int sec^3theta + int sec theta. d theta`

`I = sec theta.tan theta - I + ln  |sec theta + tan theta| + C`

`2I = sec theta.tan theta + ln  |sec theta + tan theta| + C`

`I = 1/2 [sec theta.tan theta + ln  |sec theta + tan theta| + C]`