Mathematics

# Evaluate: $\displaystyle\int {{{2 - 3\sin x} \over {{{\cos }^2}x}}dx}$

##### SOLUTION
Solution:
$I=\displaystyle \int { \frac { 2-3\sin { x } }{ \cos ^{ 2 }{ x } } dx }$

$=\displaystyle \int { 2\sec ^{ 2 }{ x } dx } -3\int { \frac { \sin { x } }{ \cos ^{ 2 }{ x } } dx }$

$\displaystyle =2\tan { x } -3\int { \frac { \sin { x } }{ \cos ^{ 2 }{ x } } dx }$

Let $\cos x=t\quad \Rightarrow \ -\sin x dx=dt$

$\displaystyle I=2\tan { x } +3\int { \frac { 1 }{ { t }^{ 2 } } dx }$

$=2\tan x-\dfrac {3}{t}+c\ \Rightarrow \ 2\tan x-\dfrac {3}{\cos x}+c=2\tan x-3\sec x+c$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 128

#### Realted Questions

Q1 Subjective Medium
$\text { Evaluate: } \displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 2 \theta}{\sin ^{4} \theta+\cos ^{4} \theta} \mathrm{d} \theta$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Find :-
$\int {\left[ {\log (\log x) + \frac{1}{{(\log {x^2})}}} \right]dx}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int\frac{dx}{x(x^7+1)}$ is equal to
• A. $log\left(\dfrac{x^7}{x^7+1}\right)+c$
• B. $log\left(\dfrac{x^7+1}{x^7}\right)+c$
• C. $\dfrac{1}{7}log\left(\dfrac{x^7+1}{x^7}\right)+c$
• D. $\dfrac{1}{7}log\left(\dfrac{x^7}{x^7+1}\right)+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle \int _{0}^{1}\frac{\sin t}{1+t} dt = \alpha$. then value of
$\displaystyle I=\int_{4\pi-2}^{4\pi}\frac{\sin(x/2)}{4\pi+2-x} dx$
• A. $\alpha /2$
• B. $-\alpha/2$
• C. $\alpha$
• D. $-\alpha$

Consider the integrals $I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tan x}}$ and $I_2=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{sin x}dx}{\sqrt{sin }x+\sqrt{cos}x}$