Answer to Question #204919 in Calculus for Visvjeet Atpadkar

Question

) If 𝐼𝑛 = ∫ 𝑐𝑜𝑠𝑛𝑥 cos 𝑛𝑥 𝑑𝑥

𝜋

2

then find the reduction formula 

connecting 𝐼𝑛 and 𝐼𝑛−1 and hence prove that 𝐼𝑛 =

𝜋

2

𝑛+1

 

Answer 1

Let

$$I_{m}=\int\cos ^m(x)dx=\int\cos ^{m-1}(x)\cos(x)dx$$

Integration by parts

$$\int udv=uv-\int vdu$$

$$u=\cos^{m-1}( x), du=-(m-1)\cos^{m-2}(x)\cdot\sin xdx$$

$$dv=\cos(x)dx, v=\sin(x)$$

$$I_{m}=\int\cos ^m(x)dx=\int\cos ^{m-1}(x)\cos(x)dx$$

$$=\cos^{m-1}( x)\sin (x)+(m-1)\int\cos ^{m-2}(x)\sin^2(x)dx$$

$$=\cos^{m-1}( x)\sin (x)+(m-1)\int\cos ^{m-2}(x)dx$$

$$-(m-1)\int\cos ^{m-2}(x)\cos^2(x)dx$$

$$=\cos^{m-1}( x)\sin (x)+(m-1)I_{m-2}-(m-1)I_{m}, m\geq 2$$

Solve for $I_{m}$

$$I_{m}=\dfrac{\cos^{m-1}( x)\sin (x)}{m}+\dfrac{m-1}{m}I_{m-2}$$

Theerefore

$$I_n=\displaystyle\int_{0}^{\pi/2}\cos ^n(x)dx$$

$$=\bigg[\dfrac{\cos^{n-1}( x)\sin (x)}{n}\bigg]\begin{matrix} \pi/2 \\ 0 \end{matrix}+\dfrac{n-1}{n}I_{n-2}$$

$$=\dfrac{n-1}{n}I_{n-2}=\dfrac{n-1}{n}\displaystyle\int_{0}^{\pi/2}\cos ^{n-2}(x)dx, n\geq2$$

$$I_n=\displaystyle\int_{0}^{\pi/2}\cos ^n(x)dx=\dfrac{n-1}{n}\displaystyle\int_{0}^{\pi/2}\cos ^{n-2}(x)dx$$

$$=\dfrac{n-1}{n}I_{n-2}, n\geq 2$$

$$n=0: I_0=\displaystyle\int_{0}^{\pi/2}\cos ^0(x)dx=[x]\begin{matrix} \pi/2 \\ 0 \end{matrix}=\dfrac{\pi}{2}$$

$$n=1: I_0=\displaystyle\int_{0}^{\pi/2}\cos ^1(x)dx=[\sin(x)]\begin{matrix} \pi/2 \\ 0 \end{matrix}=1$$