The integral I= displaystyle ∫ sec3 x tan3 ⁡ x d x is equal to (where, C is the constant of integration)

Question

The integral I= displaystyle ∫ sec3 x tan3 ⁡ x d x is equal to (where, C is the constant of integration) (A) sec5 x - sec3 ⁡ x + C (B) (sec5 x/5)-sec3 â¡ x + C (C) (sec5 x/5)-(sec3 â¡ x/3)+C (D) (sec5 x/5)-tan- 1 â¡ x+C

Tardigrade Admin · Accepted Answer

Let sec x = t⇒ sec â¡ xtan â¡ xdx=dt ∴ I= displaystyle ∫ (sec)2 x(tan)2â¡x(sec x tan x d x) = displaystyle ∫ t2(t2 - 1)dt = displaystyle ∫ (t4 - t2)dt =(t5/5)-(t3/3)+C I=(sec5 x/5)-(sec3 â¡ x/3)+C

Q. The integral $I = \int se c^{3} x t a n^{3} x d x$ is equal to (where, $C$ is the constant of integration)

$se c^{5} x - se c^{3} x + C$

$\frac{se c ^{5} x}{5} - se c^{3} x + C$

$\frac{se c ^{5} x}{5} - \frac{se c ^{3} x}{3} + C$

$\frac{se c ^{5} x}{5} - t a n^{- 1} x + C$

Let $sec x = t \Rightarrow sec x t an x d x = d t$
$∴ I = \int (sec)^{2} x (t an)^{2} x (sec x t an x d x)$
$= \int t^{2} (t^{2} - 1) d t$
$= \int (t^{4} - t^{2}) d t$
$= \frac{t ^{5}}{5} - \frac{t ^{3}}{3} + C$
$I = \frac{se c ^{5} x}{5} - \frac{se c ^{3} x}{3} + C$

Q. The integral I=∫sec3xtan3⁡xdx is equal to (where, C is the constant of integration)

sec5x−sec3⁡x+C

5sec5x​−sec3⁡x+C

5sec5x​−3sec3⁡x​+C

5sec5x​−tan−1⁡x+C