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Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 56 maths.

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Answer:   \log \frac{9}{8}

Hint: Use indefinite integral formula and the limits to solve this integral

Given: \int_{0}^{\frac{\pi}{2}}\frac{\sin x.\cos x}{\cos^2x+3\cos x+2}dx


\int_{0}^{\frac{\pi}{2}}\frac{\sin x.\cos x}{\cos^2x+3\cos x+2}dx

\begin{aligned} &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t\\ &\text { Now } x=0 \text { then } \mathrm{t}=1 \& \text { when } x=\frac{\pi}{2} \text { then } \mathrm{t}=0\\ &\therefore \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\cos ^{2} x+3 \cos x+2} d x=\int_{1}^{0} \frac{t}{t^{2}+3 t+2}(-d t)^{\mid}\\ &=-\int_{1}^{0} \frac{t}{t^{2}+3 t+2} d t=-\int_{1}^{0} \frac{t}{(t+1)(t+2)} d t\\ &=\int_{0}^{1} \frac{t}{(t+1)(t+2)} d t \end{aligned}

To solve this integral, first we need to find it’s partial fraction then integrate it by using indefinite formula.


\begin{aligned} &\Rightarrow t=\frac{A(t+1)(t+2)}{(t+1)}+\frac{B(t+1)(t+2)}{(t+2)} \\ &\Rightarrow t=A(t+2)+B(t+1) \\ &\Rightarrow t=A t+2 A+B t+B=(A+B) t+(2 A+B) \end{aligned}

Equating coefficient of t and constant resp. , then
1=A+B                                                     ...a
0=2A+B⇒B=-2A                                      ...b
Put the value of B in a
From a⇒A+B=1⇒-1+B=1⇒B=1+1=2
\begin{aligned} &\frac{t}{(t+1)(t+2)}=\frac{-1}{(t+1)}+\frac{2}{(t+2)}\\ &\therefore \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\cos ^{2} x+3 \cos x+2} d x=\int_{0}^{1} \frac{-1}{t+1} d t+\int_{0}^{1} \frac{2}{t+2} d t\; \; \; \; \; \; \; \; \; \; ..........(i) \end{aligned}

\begin{aligned} &=-[\log (t+1)]_{0}^{1}+2[\log (t+2)]_{0}^{1} \\ &=-[\log 2-\log 1]+2[\log 3-\log 2] \\ &=-\log 2+0+2 \log 3-2 \log 2 \\ &=2 \log 3-3 \log 2 \\ &=\log 3^{2}-\log 2^{3} \\ &=\log 9-\log 8 \\ &=\log \frac{9}{8} \end{aligned}

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