Question

$\int {\sin }^{7}xdx$

Answer 1

∫ sin⁷ x dx
= ∫ sin⁶ x . sin x dx
= ∫ (sin² x)³ sin x dx
= ​​∫ (1 – cos² x)³ sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos² x)³ sin x dx
= ∫ (1 – t²)³ . (–dt)
= –​∫ (1 – t⁶ – 3t² + 3t⁴) dt
$$\begin{gathered} =-\left[t-\frac{t^7}{7}-t^3+\frac{3t^5}{5}\right]+C \\ =-\left[\cos x-\frac{{\cos }^{7}x}{7}-{\cos }^{3}x+\frac{3}{5}{\cos }^{5}x\right]+C \\ =-\cos x+\frac{1}{7}{\cos }^{7}x+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+C \end{gathered}$$