The integral \int\limits_0^\frac\pi2 \frac13+2 \sin x+\cos x ~d x is equal to :

Definite Integration PYQ — JEE Main 2022 · 29 Jul · Shift 1 · Q30 with Solution

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_{0}^{\pi/2}\frac{dx}{3+2\sin x+\cos x}.$
Use the tangent half-angle substitution: $t=\tan\frac{x}{2},\qquad dx=\frac{2\,dt}{1+t^2},$ with $\sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2}.$

When $x=0$ , $t=0$ ; and when $x=\pi/2$ , $t=\tan(\pi/4)=1$ .

Substitute into the denominator: $3+2\sin x+\cos x=3+\frac{4t}{1+t^2}+\frac{1-t^2}{1+t^2}.$ Bring to a common denominator:

=\frac{2t^2+4t+4}{1+t^2} =\frac{2(t^2+2t+2)}{1+t^2}.$$ Thus, $$I=\int_0^1 \frac{1}{\frac{2(t^2+2t+2)}{1+t^2}}\cdot \frac{2\,dt}{1+t^2}$$ so the factors simplify nicely: $$I=\int_0^1 \frac{dt}{t^2+2t+2}.$$ 4. Complete the square: $$t^2+2t+2=(t+1)^2+1.$$ Hence, $$I=\int_0^1 \frac{dt}{(t+1)^2+1}.$$ 5. Integrate: $$\int \frac{dt}{(t+1)^2+1}=\tan^{-1}(t+1).$$ Therefore, $$I=\left[\tan^{-1}(t+1)\right]_0^1 =\tan^{-1}(2)-\tan^{-1}(1).$$ Since $$\tan^{-1}(1)=\frac{\pi}{4},$$ we get $$I=\tan^{-1}(2)-\frac{\pi}{4}.$$ 6. Match with the options: - A: $\tan^{-1}(2)$ - B: $\tan^{-1}(2)-\frac{\pi}{4}$ - C: $\frac12\tan^{-1}(2)-\frac{\pi}{8}$ - D: $\frac12$ So the correct option is **B**.

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_{0}^{\pi/2}\frac{dx}{3+2\sin x+\cos x}.$
Use the tangent half-angle substitution: $t=\tan\frac{x}{2},\qquad dx=\frac{2\,dt}{1+t^2},$ with $\sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2}.$

When $x=0$ , $t=0$ ; and when $x=\pi/2$ , $t=\tan(\pi/4)=1$ .

Substitute into the denominator: $3+2\sin x+\cos x=3+\frac{4t}{1+t^2}+\frac{1-t^2}{1+t^2}.$ Bring to a common denominator:

=\frac{2t^2+4t+4}{1+t^2} =\frac{2(t^2+2t+2)}{1+t^2}.$$ Thus, $$I=\int_0^1 \frac{1}{\frac{2(t^2+2t+2)}{1+t^2}}\cdot \frac{2\,dt}{1+t^2}$$ so the factors simplify nicely: $$I=\int_0^1 \frac{dt}{t^2+2t+2}.$$ 4. Complete the square: $$t^2+2t+2=(t+1)^2+1.$$ Hence, $$I=\int_0^1 \frac{dt}{(t+1)^2+1}.$$ 5. Integrate: $$\int \frac{dt}{(t+1)^2+1}=\tan^{-1}(t+1).$$ Therefore, $$I=\left[\tan^{-1}(t+1)\right]_0^1 =\tan^{-1}(2)-\tan^{-1}(1).$$ Since $$\tan^{-1}(1)=\frac{\pi}{4},$$ we get $$I=\tan^{-1}(2)-\frac{\pi}{4}.$$ 6. Match with the options: - A: $\tan^{-1}(2)$ - B: $\tan^{-1}(2)-\frac{\pi}{4}$ - C: $\frac12\tan^{-1}(2)-\frac{\pi}{8}$ - D: $\frac12$ So the correct option is **B**.