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Question: Evaluate the value of \(\int {{{\sin }^5}xdx} \)....

Evaluate the value of ∫sin⁡5xdx\int {{{\sin }^5}xdx} .

Explanation

Solution

Here, we are asked to find the value of ∫sin⁡5xdx\int {{{\sin }^5}xdx} .
Let I=∫sin⁡5xdxI = \int {{{\sin }^5}xdx} and write sin⁡5x{\sin ^5}x as sin⁡4x×sin⁡x{\sin ^4}x \times \sin x.
Then, sin⁡4x{\sin ^4}x can be written as (sin⁡2x)2{\left( {{{\sin }^2}x} \right)^2} and sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x.
Thus, solve the question further to get the required answer.

Complete step by step solution:
Here, we are asked to find the value of ∫sin⁡5xdx\int {{{\sin }^5}xdx} .
Let I=∫sin⁡5xdxI = \int {{{\sin }^5}xdx} .
Now, we can write sin⁡5x{\sin ^5}x as sin⁡4x×sin⁡x{\sin ^4}x \times \sin x .
∴I=∫sin⁡4x×sin⁡xdx\therefore I = \int {{{\sin }^4}x \times \sin xdx}
Also, sin⁡4x{\sin ^4}x can be written as (sin⁡2x)2{\left( {{{\sin }^2}x} \right)^2} .
∴I=∫(sin⁡2x)2sin⁡xdx\therefore I = \int {{{\left( {{{\sin }^2}x} \right)}^2}\sin xdx}
Applying the property sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x in the above value of I.
∴I=∫(1−cos⁡2x)2sin⁡xdx\therefore I = \int {{{\left( {1 - {{\cos }^2}x} \right)}^2}\sin xdx}
Now, to solve further, let cos⁡x=t\cos x = t
∴−sin⁡xdx=dt ∴sin⁡xdx=−dt  \therefore - \sin xdx = dt \\\ \therefore \sin xdx = - dt \\\
∴I=−∫(1−t2)2dt ∴I=−∫(1−2t2+t4)dt ∴I=−∫dt+2∫t2dt−∫t4dt ∴I=−t+2t33−t55+C ∴I=−t55−2t33−t+C  \therefore I = - \int {{{\left( {1 - {t^2}} \right)}^2}dt} \\\ \therefore I = - \int {\left( {1 - 2{t^2} + {t^4}} \right)dt} \\\ \therefore I = - \int {dt} + 2\int {{t^2}dt} - \int {{t^4}dt} \\\ \therefore I = - t + 2\dfrac{{{t^3}}}{3} - \dfrac{{{t^5}}}{5} + C \\\ \therefore I = - \dfrac{{{t^5}}}{5} - \dfrac{{2{t^3}}}{3} - t + C \\\
Now, returning back the value of t as cos⁡x\cos x .

∴I=−cos⁡5x5+2cos⁡3x3−cos⁡x+C\therefore I = - \dfrac{{{{\cos }^5}x}}{5} + \dfrac{{2{{\cos }^3}x}}{3} - \cos x + C
Thus, ∫sin⁡5xdx=−cos⁡5x5+2cos⁡3x3−cos⁡x+C\int {{{\sin }^5}xdx} = - \dfrac{{{{\cos }^5}x}}{5} + \dfrac{{2{{\cos }^3}x}}{3} - \cos x + C.

Note:
Some properties of definite integration:

  1. ∫abf(x)dx=∫ab(a+b−f(x))dx\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {\left( {a + b - f\left( x \right)} \right)dx} }
  2. ∫−aaf(x)dx=−∫a−af(x)dx\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_a^{ - a} {f\left( x \right)dx}
  3. ∫abkxdx=k∫abxdx\int\limits_a^b {kxdx = } k\int\limits_a^b {xdx}
  4. ∫aaf(x)dx=0\int\limits_a^a {f\left( x \right)dx = } 0
    ∫abf(x)dx=−∫baf(x)dx\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx}