RD Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.7 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.7
CategoryRD Sharma Solutions

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 dxm≠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 dxm≠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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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