Mathematics

Evaluate:$$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\sin}^{3}{x}\,dx}$$


SOLUTION
$$\displaystyle\int_{0}^{\frac{\pi}{2}}{{\sin}^{3}{x}\,dx}$$
$$=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{4{\sin}^{3}{x}\,dx}$$
We know that $$\sin{3x}=3\sin{x}-4{\sin}^{3}{x}\Rightarrow\,4{\sin}^{3}{x}=3\sin{x}-\sin{3x}$$
$$=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(3\sin{x}-\sin{3x}\right)dx}$$
$$=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{3\sin{x}\,dx}-\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{3x}\,dx}$$
$$=\dfrac{3}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{x}\,dx}-\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{3x}\,dx}$$
$$=\dfrac{3}{4}\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}-\dfrac{1}{4}\left[\dfrac{-\cos{3x}}{3}\right]_{0}^{\frac{\pi}{2}}$$
$$=\dfrac{-3}{4}\left[\cos{\dfrac{\pi}{2}}-\cos{0}\right]+\dfrac{1}{12}\left[\cos{\dfrac{3\pi}{2}}-\cos{0}\right]$$
$$=\dfrac{-3}{4}\left[0-1\right]+\dfrac{1}{12}\left[\cos{\left(\pi+\dfrac{\pi}{2}\right)}-1\right]$$
$$=\dfrac{3}{4}+\dfrac{1}{12}\left[\cos{\dfrac{\pi}{2}}-1\right]$$
$$=\dfrac{3}{4}+\dfrac{1}{12}\left[0-1\right]$$
$$=\dfrac{3}{4}-\dfrac{1}{12}$$
$$=\dfrac{3\times 3}{4\times 3}-\dfrac{1}{12}$$
$$=\dfrac{9}{12}-\dfrac{1}{12}$$
$$=\dfrac{8}{12}=\dfrac{2}{3}$$

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
If $$\displaystyle \int { \left( u\cfrac { dv }{ dx }  \right)  } dx=uv-\int { wdx } $$, then $$w=$$
  • A. $$\cfrac { du }{ dx } \cfrac { dv }{ dx } $$
  • B. $$\cfrac { d }{ dx } (uv) $$
  • C. $$u\cfrac { dv }{ dx } $$
  • D. $$v\cfrac { du }{ dx } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
Let $$F\left( x \right) =f\left( x \right) +f\left( \dfrac { 1 }{ x }  \right) $$, where $$\displaystyle f\left( x \right) =\int _{ 1 }^{ x }{ \dfrac { \log { t }  }{ 1+t }  } dt$$. Then $$F\left( e \right) $$ equals -
  • A. $$0$$
  • B. $$1$$
  • C. $$2$$
  • D. $$\dfrac { 1 }{ 2 } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Prove that $$\displaystyle \int _{-a}^{a} x^{3}\sqrt{a^{2}-x^{2}}dx=0$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Medium
Evaluate $$ \int x^6 dx  $$
  • A. $$ 7x^7 +C $$
  • B. $$ 6x^2 +C $$
  • C. $$ 6x^7 +C $$
  • D. $$ \dfrac {x^7}{7} +C $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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 $$\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer