Mathematics

# Evaluate: $\displaystyle\int \dfrac{ dx} {{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}$

##### SOLUTION
$I=\int \dfrac{1}{\cos^{2} x (1- \tan x)^2} dx$

$=\int \dfrac{\sec^{2}x}{(1-\tan x)^{2}} dx$

Let $\tan x=t \Rightarrow \sec^{2} xdx=dt$

$I=\int \dfrac{dt}{(1-t)^{2}}$

$I=\dfrac{1}{(1-t)}+C$

$I=\dfrac{1}{(1-\tan x)}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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#### Realted Questions

Q1 Multiple Correct Hard
If $\displaystyle \int \frac {xe^x}{\sqrt {1 + e^x}} dx = f(x) \sqrt {1 + e^x} -2 \log \: g(x) + C$, then
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• D. $\displaystyle f(x) = 2(x - 2)$

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Q5 Passage Hard

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$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

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Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020