∫ tan 3 2 x sec 2 x d x is equal to :

Question

∫ tan 3 2 x sec 2 x d x is equal to : (A) (1/3) sec 3 2 x-(1/2) sec 2 x+c (B) -(1/6) sec 3 2 x-(1/2) sec 2 x+c (C) (1/6) sec 3 2 x-(1/2) sec 2 x+c (D) (1/3) sec 3 2 x+(1/2) sec 2 x+c

Tardigrade Admin · Accepted Answer

∫ tan 3 2 x ⋅ sec 2 x d x I=∫( sec 2 2 x-1) sec 2 x ⋅ tan 2 x d x Put sec 2 x=t ⇒ 2 sec 2 x tan 2 x d x=d t ⇒ I=(1/2) ∫(t2-1) d t =(t3/6)-(t/2)+C=(1/6) sec 3 2 x-(1/2) sec 2 x+C

Q. $\int tan^{3} 2 x sec 2 x d x$ is equal to :

$\frac{1}{3} sec^{3} 2 x - \frac{1}{2} sec 2 x + c$

$- \frac{1}{6} sec^{3} 2 x - \frac{1}{2} sec 2 x + c$

$\frac{1}{6} sec^{3} 2 x - \frac{1}{2} sec 2 x + c$

$\frac{1}{3} sec^{3} 2 x + \frac{1}{2} sec 2 x + c$

$\int tan^{3} 2 x \cdot sec 2 x d x$
$I = \int (sec^{2} 2 x - 1) sec 2 x \cdot tan 2 x d x$
Put $sec 2 x = t \Rightarrow 2 sec 2 x tan 2 x d x = d t$
$\Rightarrow I = \frac{1}{2} \int (t^{2} - 1) d t$
$= \frac{t ^{3}}{6} - \frac{t}{2} + C = \frac{1}{6} sec^{3} 2 x - \frac{1}{2} sec 2 x + C$

Q. ∫tan32xsec2xdx is equal to :

31​sec32x−21​sec2x+c

−61​sec32x−21​sec2x+c

61​sec32x−21​sec2x+c

31​sec32x+21​sec2x+c