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  • Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 16 maths

Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 16 maths

Answers (1)

Answer:

\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan ^{-1}(\sin x+\cos x)+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

\int \frac{1}{\cos x+\operatorname{cosec} x} d x

Solution:

Let\: \: I=\int \frac{1}{\cos x+\operatorname{cosec} x} d x

\begin{aligned} &=\frac{1}{\cos x+\frac{1}{\sin x}} d x \quad\left[\because \operatorname{cosec} x=\frac{1}{\sin x}\right] \\ &=\int \frac{1}{\frac{\cos x \cdot \sin x+1}{\sin x}} d x=\int \frac{\sin x}{\cos x \cdot \sin x+1} d x \\ &=\int \frac{2 \sin x}{2(\cos x \cdot \sin x+1)} d x=\int \frac{\sin x+\sin x}{2+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x-\cos x+\sin x}{2+2 \sin x \cdot \cos x} d x \\ &=\int\left\{\frac{\sin x+\cos x}{2+2 \sin x \cdot \cos x}+\frac{\sin x-\cos x}{2+2 \sin x \cdot \cos x}\right\} d x \end{aligned}

\begin{aligned} &=\int \frac{\sin x+\cos x}{2+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{2+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x}{3-1+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+1+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x}{3-\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+\left(\sin x+\cos ^{2} x\right)+2 \sin x \cdot \cos x} d x \\ &{\left[\because 1=\sin ^{2} x+\cos ^{2} x\right]} \\ &=\int \frac{\sin x+\cos x}{3-\sin ^{2} x-\cos ^{2} x+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cdot \cos x\right)} d x \end{aligned}

\begin{aligned} &=\int \frac{\sin x+\cos x}{3-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cdot \cos x\right)} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &\text { Let } I=I_{1}+I_{2}\: \: \: \: \: \: ...(i) \\ &\Rightarrow \int \frac{1}{\cos x+\operatorname{cosec} x} d x=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &\text { Where } I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x \end{aligned}

\text { And } I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x

Here first we find the integral of I1 and I2 then we put the value of  I1 and I2  in I  and then we get the required solution

\begin{aligned} &\text { So } I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x \\ &\text { Put } t=\sin x-\cos x \Rightarrow d t=[\cos x-(-\sin x)] d x \Rightarrow d t=[\cos x+\sin x] d x \\ &\text { Then } I_{1}=\int \frac{1}{3-t^{2}} d t=\int \frac{1}{(\sqrt{3})^{2}-t^{2}} d t \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C_{1} \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-(\sin x-\cos x)}\right|+C_{1} \quad[\because t=\sin x-\cos x] \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|+C_{1}\: \: \: \: ......(ii) \\ &\text { And } I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \end{aligned}

\begin{aligned} &\text { Put } u=\sin x+\cos x \Rightarrow d u=(\cos x-\sin x) d x \Rightarrow d u=-(\sin x-\cos x) d x\\ &\Rightarrow(\sin x-\cos x) d x=-d u\\ &\therefore I_{2}=\int \frac{1}{1+u^{2}}(-d u)=-\int \frac{1}{1^{2}+u^{2}} d u\\ &=-\tan ^{-1}\left(\frac{u}{1}\right)+C_{2} \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]\\ &=-\tan ^{-1}(\sin x+\cos x)+C_{2}\: \: \: \: .....\text { (iii) }[\because u=\sin x+\cos x]\\ &\text { Putting the value of (ii) and (iii) in (i) we get }\\ &I=I_{1}+I_{2}\\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|+C_{1}+\left\{-\tan ^{-1}(\sin x+\cos x)+C_{2}\right\}\\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan ^{-1}(\sin x+\cos x)+C \quad\left[\because C=C_{1}+C_{2}\right] \end{aligned}

Posted by

Gurleen Kaur

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