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Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.7

  • Last Updated : 13 Jan, 2021

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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