Mathematics

# $\displaystyle \int \dfrac{\sec^2 x}{\sec x+\tan x)^5}dx=$

##### SOLUTION
$\displaystyle \int \frac{\sec^{2}x\ dx}{(\sec x+\tan x)^{5}}$
Take $\sec x$ out in denominator
$\displaystyle =\int \frac{\sec^{2}x \ dx}{\sec^{5}x (1+ \sin x)^{5}}$

$\displaystyle =\int \frac{\cos^{3}x \ dx}{(1+\sin x)^{5}}$
Put $1+ \sin x=z$
$\cos x \ dx = dz$
$\displaystyle =\int \frac{1-z^{2}}{z^{5}}dz$
$\displaystyle =\int \frac{1}{z^{5}}-\frac{1}{z^{3}}dz$
$\displaystyle =-\frac{1}{4{z^{4}}}+\frac{1}{2{z^{2}}}+C$
$\displaystyle =-\frac{1}{4(\sin x+1)^{4}}+\frac{1}{2(\sin x+1)^{2}}+C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\overset { 4 }{ \underset { -4 }{ \int } }$ log $\left(\dfrac{9 - x}{9 + x}\right)$ dx equals.
• A. -4
• B. 8
• C.
• D. 4

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int (27e^{9x} + e^{12x})^{1/3}dx$ is equal to
• A. $(1/4)(27 + e^{3x})^{1/3} + C$
• B. $(1/4)(27 + e^{3x})^{2/3} + C$
• C. $(1/3)(27 + e^{3x})^{4/3} + C$
• D. $\dfrac {3}{4}(27 + e^{3x})^{4/3} + C$
• E. $(1/4)(27 + e^{3x})^{4/3} + C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int \frac{\sec x dx}{\sqrt{\sin(2x+A)+\sin A}}$ is equal to
• A. $\displaystyle \frac{\sec A}{\sqrt{2}}\sqrt{\tan x\cos A-\sin A}+C$
• B. $\displaystyle \sqrt{2} \sec A\sqrt{\tan x\cos A-\sin A}+C$
• C. none of these
• D. $\displaystyle \sqrt{2} \sec A\sqrt{\tan x\cos A+\sin A}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Hard
*$\displaystyle\int \sqrt{\left ( \frac{x}{a^{3}-x^{3}} \right )\cdot }dx=\frac{6}{k}\sin ^{-1}\frac{x^{3/2}}{a^{3/2}}.$ Find the value of $k$.

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$