∫ ( tan 4 √x ⋅ sec 2 √x/√x) d x=

Question

∫ ( tan 4 √x ⋅ sec 2 √x/√x) d x= (A) (-5/2)[ tan √x]5+C (B) [ tan √x]5+C (C) (2/5)[ tan √x]5+C (D) (5/2)[ tan √x]5+C

Tardigrade Admin · Accepted Answer

Let I=∫ ( tan 4 √x ⋅ sec 2 √x/√x) d x Put tan √x=t ⇒ ( sec 2 √x/2 √x) d x=d t ∴ I=∫ ( tan 4 √x ⋅ sec 2 √x/√x) d x=∫ 2 t4 d t =(2 t5/5)+C=(2( tan √x)5/5)+C

Q. $\int \frac{t a n ^{4} x \cdot s e c ^{2} x}{x} d x =$

$\frac{- 5}{2} [tan x]^{5} + C$

$[tan x]^{5} + C$

$\frac{2}{5} [tan x]^{5} + C$

$\frac{5}{2} [tan x]^{5} + C$

Let $I = \int \frac{t a n ^{4} x \cdot s e c ^{2} x}{x} d x$
Put $tan x = t$
$\Rightarrow \frac{s e c ^{2} x}{2 x} d x = d t$
$∴ I = \int \frac{t a n ^{4} x \cdot s e c ^{2} x}{x} d x = \int 2 t^{4} d t$
$= \frac{2 t ^{5}}{5} + C = \frac{2 ( t a n x ) ^{5}}{5} + C$

Q. ∫x​tan4x​⋅sec2x​​dx=

2−5​[tanx​]5+C

[tanx​]5+C

52​[tanx​]5+C

25​[tanx​]5+C