Mathematics

# Evaluate $I = \displaystyle \int_{\pi /6}^{\pi /3}\sin x\:dx$

$\displaystyle \frac{1-\sqrt{3}}{2}$

##### SOLUTION
Given : $I = \displaystyle \int_{\pi /6}^{\pi /3}\sin x\:dx$

Integeration of $\sin x dx$ is $-cos x dx + c$

$I = -cos x dx$

Substuting the upper and lower limit values we get,

$I = -cos\dfrac{\pi}{3}+cos\dfrac{\pi}{6}$

$I = \dfrac{-\sqrt{3}}{2} + \dfrac{1}{2}$

$I = \dfrac{1-\sqrt{3}}{2}$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\frac{(\log x)^{2}-\log x+1}{[(\log x)^{2}+1]^{3/2}}dx=$
• A. $\displaystyle \frac{x}{[(\log{x^{2}}+1]^{3/2}}+c$
• B. $\displaystyle \frac{x+1}{\sqrt{(\log x)^{2}}+1}+c$
• C. $\displaystyle \frac{x}{[(\log x)^{2}+1]^{3/2}}+c$
• D. $\displaystyle \frac{x}{\sqrt{(\log x)^{2}+1}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve $\displaystyle\int {\dfrac{{{{\cos }^2}\theta d\theta }}{{{{\cos }^2}\theta + 4{{\sin }^2}\theta }}}$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Integrate w.r.t $x$:$\dfrac{dx}{{\left({\sin}^{3}{x}{\cos}^{5}{x}\right)}^{\frac{1}{4}}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\displaystyle \int { \cos{ \pi }xdx=? }$

$\int \frac{x}{x^2 + a^2} \;dx$