Mathematics

# Solve : $\displaystyle \int \dfrac{\sin x \cos^3 x}{1 + \cos^2 \ x} dx$

##### SOLUTION
$\displaystyle \int \dfrac{\sin x\ \cos^3x}{1+\cos^2x}dx$

$\cos x=p$

$-\ sinx dx=dp$

$\Rightarrow \displaystyle -\int \dfrac{p^3}{1+p^2}dp$

$1+p^2=k \Rightarrow p^2=(k-1)$

$2pdp=dk$

$\Rightarrow \displaystyle \int \dfrac{(k-1)dk}{2(k)}$

$\therefore \displaystyle \int \dfrac{1}{2}dx+\dfrac{1}{2} \int \dfrac{dx}{k}$

$\therefore \displaystyle \int -\dfrac{k}{2}+\dfrac{1}{2}ln|k|$

$\therefore \dfrac{-(1+p^2)}{2}+\dfrac{1}{2}ln|1+p^2|$

$\therefore \boxed{\dfrac{1}{2}\left[ln|1+\cos^2x)-(1+\cos^2x)\right]}$

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

#### Realted Questions

Q1 Single Correct Medium
The value of the integral $\displaystyle \int_0^1 \dfrac{x^3}{1 + x^3} dx$ is equal to
• A. $\dfrac{\pi}{8}$
• B. $\dfrac{\pi}{4}$
• C. $\dfrac{\pi}{6}$
• D. $\dfrac{\pi}{12}$
• E. $\dfrac{\pi}{16}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium

$\displaystyle \int_{\pi^{2}/16}^{\pi^{2}/4}\frac{\sin\sqrt{x}}{\sqrt{x}}dx=$
• A. $1/\sqrt{2}$
• B. 2 $\sqrt{2}$
• C. ${\pi}/2$
• D. $\sqrt{2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
integrate :
$\int {{{\tan }^{ - 1}}xdx} .$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value of the integral $\displaystyle \int \frac{(1-\cos\theta )^{2/7}}{(1+\cos\theta )^{9/7}}d\theta$ is
• A. $\dfrac{7}{11}\left (\cos\dfrac{\theta }{2}\right)^{\dfrac{11}{7}}+C$
• B. $\dfrac{7}{11}\left (\sin\dfrac{\theta }{2}\right)^{\dfrac{11}{7}}+C$
• C. None of these
• D. $\displaystyle \frac{7}{11}\left (\tan\frac{\theta }{2}\right)^{\dfrac{11}{7}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$