Question

Integral of tan^3 2x sec 2x

Answer 1

Write tan³2x=tan²2x*tan2x
⇒(sec²2x-1)*tan2x
Now the integral is:∫(sec²2x-1)*tan2x.sec2x*dx
let sec2x=t
⇒2sec2x*tan2x*dx=dt
⇒sec2x*tan2x*dx=dt/2
⇒1/2∫(t²-1)*dt
⇒1/2[t³/3-t]+c
⇒sec³2x/6-sec2x/4+c

Answer 2

$I=\dfrac{1}{6} \sec^3 2x-\dfrac{1}{2} \sec 2x+c$

Step-by-step explanation:

Let $I=\int \tan^3 2x \sec 2x \, dx$

To find, $\int \tan^3 2x \sec 2x \, dx=?$

$I=\int tan^2 2x\times tan2x \sec 2x \, dx$

$I=\int (sec^22x-1)\times tan2x \sec 2x \, dx$      .....(1)

Let $\sec 2x=t$

⇒ $2\sec 2x.\tan 2xdx=dt$ [∵$\dfrac{d(\sec x)}{dx} =\sec x\tan x$]

⇒ $\sec 2x\tan 2xdx=\dfrac{dt}{2}$

Now, equation (1) becomes, we get

⇒$I=\dfrac{1}{2} \int (t^2-1) \, dt$

⇒$I=\dfrac{1}{2} [\dfrac{t^{2+1}}{2+1} -t]+C$   .....(2)

Where, C is calles integration constant.

[∵ $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}$]

Put $\sec 2x=t$ in (2), we get

⇒$I=\dfrac{1}{6} \sec^3 2x-\dfrac{1}{2} \sec 2x+c$

Hence, $I=\dfrac{1}{6} \sec^3 2x-\dfrac{1}{2} \sec 2x+c$