Mathematics

# Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x+\cos x}{3+\sin 2x}dx$

##### SOLUTION
Let $I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+\sin 2x}\ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+1-1+\sin 2x}\ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(1-\sin 2x)} \ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(\sin ^2x+\cos ^2 x-\sin 2x)} \ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(\sin x-\cos x)^{2}}\ dx$

Let $(\sin x-\cos x)=t \Rightarrow (\cos x+\sin x)=\dfrac {dt}{dx}$

$\therefore I_t=\displaystyle\int\dfrac{1}{4-(t)^{2}}\ dt$

$\therefore I=\dfrac{1}{4}\left[\log\left|\dfrac{2+\sin x-\cos x}{2-\sin x+\cos x}\right|\right]_{0}^{\pi/4}=-\dfrac{1}{4}\log\dfrac{1}{3}=\dfrac{1}{4}\log_{e}3$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
The tangent to the graph of the function $\displaystyle y = f(x)$ at the point with abscissa x = a forms with the x-axis an angle of $\displaystyle \pi/3$ and at the point with abscissa x = b at an angle of $\displaystyle \pi/4$, then the value of the integral,
$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$ is equal to
• A. 1
• B.
• C. $\displaystyle -\sqrt 3$
• D. -1

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, }$ then $I$ equals
• A. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
• B. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 2 }\sqrt { x } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
• C. None of these
• D. $\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int_{-3\pi}^{3\pi}{\sin^{2}\theta\sin^{2} 2\theta d \theta}$ is equal to-
• A. $\dfrac{3\pi}{2}$
• B. $\dfrac{5\pi}{2}$
• C. $6\pi$
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int cos(\log x)dx=$
• A. $x[cos(\log x)-sin(\log x)]+c$
• B. $\displaystyle \frac{x}{2}[cos(\log x)-sin(\log x)]+c$
• C. $\displaystyle \frac{\log x}{2}[cosx+sinx]+c$
• D. $\dfrac{x}{2}[cos(\log x)+sin(\log x)]+c$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$