RD Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	19
Exercise	19.7
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals

Integrate the following integrals:

Question 1. ∫sin4x cos7x dx

Solution:

Let I= $\int \sin4x\cos7x\,dx$

We know,

$2\sin A \cos B= \sin(A+B)+\sin(A-B)$

Applying this formula to the given question we get

I= $\int \frac 1 2(\sin(4x+7x)+\sin(4x-7x))\,dx$

= $\int \frac 1 2(\sin11x+\sin(-3x)\,dx$

= $\int\frac 1 2 (\sin 11x -\sin3x)\,dx$ $[\because \sin(-\theta)=-\sin\theta]$

= $\frac 1 2(\int\sin 11x -\int\sin3x)\,dx$

We know,

$\int \sin ax\,dx=-\frac 1 a\cos ax+C$

Applying this formula to the given question we get

I= $\frac 1 2(-\frac {1} {11} \cos 11x +\frac 1 3 \cos3x)$

$\therefore$ I= $-\frac {1} {22} \cos 11x +\frac 1 6 \cos3x+C$

Question 2. ∫ cos3x cos4x dx

Solution:

Let I= $\int \cos3x\cos4x\,dx$

Multiplying and dividing the equation by 2,we get

I= $\frac 1 2\int 2\cos3x\cos4x\,dx$

We know,

$2\cos A \cos B= \cos(A+B)+\cos(A-B)$

Applying this formula to the given question we get

I= $\frac 1 2\int( \cos(3x+4x)+\cos(3x-4x))\,dx$

= $\frac 1 2\int( \cos 7x+\cos x)\,dx$ $[\because \cos(-\theta)=\cos\theta]$

We know,

$\int \cos x\,dx=\sin x+C$ and $\int \cos ax\,dx=\frac 1 a\sin ax+C$

Applying these formulas to the given question we get

I= $\frac 1 2 (\frac {\sin 7x} 7+\sin x) + C$

$\therefore$ I= $(\frac {\sin 7x} {14}+\frac {\sin x} 2) + C$

Question 3. ∫ cosmx cosnx dx, m≠n

Solution:

Let I= $\int \cos mx \cos nx\,dx , m \ne n$

Multiplying and dividing the equation by 2,we get

I= $\frac 1 2\int 2\cos mx\cos nx\,dx$

We know,

$2\cos A \cos B= \cos(A+B)+\cos(A-B)$

Applying this formula to the given question we get

I= $\frac 1 2\int( \cos(m+n)x+\cos(m-n)x)\,dx$

We know,

$\int \cos ax\,dx=\frac 1 a\sin ax+C$

Applying these formulas to the given question we get

$\therefore$ I= $\frac 1 2 (\frac {\sin (m+n)x} {m+n}+\frac {\sin (m-n)x} {m-n}) + C$

Question 4. ∫ sinmx cosnx dx, m≠n

Solution:

Let I= $\int \sin mx \cos nx\,dx , m \ne n$

Multiplying and dividing the equation by 2,we get

I= $\frac 1 2\int 2\sin mx\cos nx\,dx$

We know,

$2\sin A \cos B= \sin(A+B)+\sin(A-B)$

Applying this formula to the given question we get

I= $\frac 1 2\int( \sin(m+n)x+\sin(m-n)x)\,dx$

We know,

$\int \sin ax\,dx=-\frac 1 a\cos ax+C$

Applying these formulas to the given question we get

$\therefore$ I= $\frac 1 2 (-\frac {\cos (m+n)x} {m+n}-\frac {\cos (m-n)x} {m-n}) + C$

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RD Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals

Integrate the following integrals:

Question 1. ∫sin4x cos7x dx

Question 2. ∫ cos3x cos4x dx

Question 3. ∫ cosmx cosnx dx, m≠​n

Question 4. ∫ sinmx cosnx dx, m≠n

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Question 3. ∫ cosmx cosnx dx, m≠n