Mathematics

# Evaluate the following definite integrals :$\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$\frac { \pi }{ 4 }$

##### SOLUTION

$I=\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$........$using \ (\cos2 x=2cos^2 x-1)$

$=\dfrac {1}{2} \left [x +\dfrac {\sin 2x}{2} \right]_0^{\pi /2}$

$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right] =\dfrac {\pi}{4}$

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

#### Realted Questions

Q1 Subjective Medium
Let $f(x)=\left\{\begin{matrix}3x^2+4, & when & 0\leq x\leq 2\\ 9x-2, & when & 2\leq x\leq 4\end{matrix}\right.$. Show that $\displaystyle\int^4_0f(x)dx=66$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Write a value of
$\displaystyle\int { { e }^{ x }\left( \cfrac { 1 }{ x } -\cfrac { 1 }{ { x }^{ 2 } } \right) } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Assertion & Reason Hard
##### ASSERTION

If $\displaystyle f\left( x \right)=\int _{ 0 }^{ x }{ g\left( t \right) dt }$, where $g$ is an even function and $f\left( x+5 \right) =g\left( x \right)$, then $\displaystyle g\left( 0 \right)-g\left( x \right)=\int _{ 0 }^{ x }{ f\left( t \right)dt }$

##### REASON

$f$ is an odd function.

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

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate the following integrals
$\int { { x }^{ 2 }\cos { 2x } } dx$

Given that for each $\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$ exists. Let this limit be $g(a)$
In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$