Question

Can you plz solve it for me with a proper explanation?​

Answer 1

Formula Applied :

$\bullet\ \; \displaystyle \sf \int_{a}^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

Solution :

Let ,

$\displaystyle \sf I=\int _{-2}^2\bigg(x^3.cos\dfrac{x}{2}+\dfrac{1}{2}\bigg)\sqrt{4-x^2}\ dx\ ...(1)$

(-2) + 2 - x = - x

So , this becomes ,

$\displaystyle \to \sf I=\int_{-2}^2\bigg((-x)^3.cos\dfrac{(-x)}{2}+\dfrac{1}{2}\bigg)\sqrt{4-x^2}\ dx\\\\\to \sf I=\int_{-2}^2\bigg(-x^3.cos\dfrac{x}{2}+\dfrac{1}{2}\bigg)\sqrt{4-x^2}\ dx\ ...(2)$

Add (1) and (2) ,

$\\ \displaystyle \to \sf 2I=\int_{-2}^2\bigg(x^3.cos\dfrac{x}{2}+\dfrac{1}{2}\bigg)\sqrt{4-x^2}+\bigg(-x^3.cos\dfrac{x}{2}+\dfrac{1}{2}\bigg)\sqrt{4-x^2}\ dx$

$\\ \displaystyle \sf \to 2I=\int_{-2}^2 \cancel{x^3cos\dfrac{x}{2}\sqrt{4-x^2}}+\dfrac{\sqrt{4-x^2}}{2}-\cancel{x^3.cos\dfrac{x}{2}\sqrt{4-x^2}}+\dfrac{\sqrt{4-x^2}}{2}\ dx$

$\\ \to \displaystyle \sf 2I=\int_{-2}^2 \sqrt{4-x^2}\ dx$

$\\ \displaystyle \to \sf 2I=2\int_{0}^2\sqrt{4-x^2}\ dx$

Let's use substitution method ,

where  , x = 2sinθ

➠ dx = 2cosθdθ

Limits ,

x  : 0  →  2

θ  : 0  →  π/₂

$\displaystyle \sf \to 2I=2\int_0^{\frac{\pi}{2}} \sqrt{4-(2sin\theta)^2}\ 2cos\theta\ d\theta$

$\\ \displaystyle \sf \to I=\int_0^{\frac{\pi}{2}}\sqrt{4-4sin^2\theta}\ 2cos\theta\ d\theta$

$\\ \displaystyle \to \sf I=4\int_0^{\frac{\pi}{2}}\sqrt{1-sin^2\theta}\ (cos \theta)\ d\theta$

$\\ \displaystyle \to \sf I=4\int_0^{\frac{\pi}{2}}\sqrt{cos^2\theta}\ (cos \theta)\ d\theta$

$\\ \displaystyle \to \sf I=4\int_0^{\frac{\pi}{2}}cos\theta\ (cos \theta)\ d\theta$

$\\ \displaystyle \to \sf I=4\int_0^{\frac{\pi}{2}}cos^2\theta\ d\theta$

$\bullet\ \; \; \sf \blue{cos^2\theta=\dfrac{1+cos2\theta}{2}}$

$\\ \displaystyle \to \sf I=4\int_0^{\frac{\pi}{2}} \bigg(\dfrac{1+cos2\theta}{2}\bigg)d\theta$

$\\ \displaystyle \to \sf I=2\int_0^{\frac{\pi}{2}} (1+cos2\theta)\ d\theta$

$\\ \to \displaystyle \sf I=2\bigg[\theta+\dfrac{sin2\theta}{2}\bigg]_0^{\frac{\pi}{2}}$

$\displaystyle \to \sf I=2\bigg[\dfrac{\pi}{2}+\dfrac{sin2\frac{\pi}{2}}{2}\bigg]$

$\to \sf I=2\bigg[\dfrac{\pi}{2}+\dfrac{0}{2}\bigg]$

$\to \sf I=2\bigg[\dfrac{\pi}{2}\bigg]$

$\leadsto \sf I=\pi\ \; \pink{\bigstar}$