Mathematics

# Evaluate:$\displaystyle\int {\dfrac{1}{\sqrt{1-(bx+c)^2}}dx}$

##### SOLUTION
$I=\displaystyle\int \dfrac{1}{\sqrt{1-(bx+c)^2}}dx$

put $x=\left[\dfrac{\cos\theta}{b}-c\right]$

$\Rightarrow \cos\theta =bx+c$

$\theta =\cos^{-1}(bx+c)$

$dx=\dfrac{-\sin\theta}{b}d\theta$

$I=\displaystyle\int \dfrac{-\sin\theta d\theta}{b\sqrt{1-\cos^2\theta}}$

$I=\displaystyle\int \dfrac{-\sin\theta}{b\sin\theta}d\theta$

$I=-\dfrac{1}{b}\displaystyle\int d\theta =-\dfrac{1}{b}\theta +k$

$\therefore \displaystyle\int \dfrac{1}{\sqrt{1-(bx+c)^2}}=-\dfrac{1}{b}\cos^{-1}(bx+c)+k$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Assertion & Reason Hard
##### ASSERTION

The value of $\displaystyle \int_{0}^{\pi /2}\sin ^{6}xdx=\frac{5\pi }{16}$

##### REASON

If n is even, then $\int_{0}^{\pi /2}\sin ^{n}xdx$ equals $\displaystyle \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}...\frac{1}{2}\times \frac{\pi }{2}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct.

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve:
$\int {\dfrac{x}{{{x^2} + x + 1}}} dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int { { e }^{ \tan ^{ -1 }{ { x }^{ } } } } \left( \cfrac { 1+x+{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) dx$ is equal to
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• C. None of the above
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard

Integrate
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$I=\displaystyle \int {\dfrac {1}{3\sin x-4\cos x}}dx$