# Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.12

### Question 1. ∫sin^{4}x cos^{3}x dx

**Solution: **

Let I = ∫ sin

^{4}x cos^{3}x dx -(i)Let sinx = t

On differentiating with respect to x:

cosx = dt/dx

cosx dx = dt

dx = dt/cosx

Putting value of dx and sinx in equation (i):

I = ∫ t

^{4 }cos^{x}dt/cosx

I = ∫ t^{4 }cos^{2 }x dtI = ∫ t

^{4 }(1 – sin^{2 }x) dtI = ∫ t

^{4 }(1 – t^{2}) dtI = ∫ (t

^{4}– t^{2}) dtI = t

^{5}/5 – t^{7}/7 + cI = sin

^{5}/5 – sin^{7}/7 + c

### Question 2. ∫ sin^{5}x dx

**Solution: **

Let I = ∫ sin

^{5}x dxI = ∫sin

^{3}xsin^{2}x dx= ∫sin

^{3}x(1 – cos^{2}x)dx= ∫(sin

^{3}x – sin^{3}xcos^{2}x)dx= ∫[sinxsin

^{2}x – sin^{3}xcos^{2}x]dx= ∫[sinx(1 – cos

^{2}x) – sin^{3}xcos^{2}x]dx= ∫(sinx – sinxcos

^{2}x – sin^{3}xcos^{2}x)dxI = ∫sinx dx – ∫sinxcos

^{2}x dx – ∫sin^{3}xcos^{2}x dxPutting cosx = t and -sinxdx = dt in 2nd and 3rd integral:

I = ∫sinx dx + ∫t

^{2}dt + ∫sin^{2}xt^{3}dt/t= ∫sinx dx + ∫t

^{2 }dt + ∫sin^{2}xt^{2 }dt= ∫sinx dx + ∫t

^{2 }dt + ∫(1 – cos^{2}x)t^{2 }dtPutting value of t:

### Question 3.** **∫cos^{5}x dx

**Solution:**

Let I = ∫cos

^{5}x dxI = ∫cos

^{2}xcos^{3}x dx= ∫(1 – sin

^{2}x)cos^{3}x dx= ∫(cos

^{3}x−sin^{2}xcos^{3}x)dx= ∫(cos

^{2}xcosx – sin^{2}xcos^{2}xcosx)dx= ∫[(1 – sin

^{2}x)cosx – sin^{2}x(1 – sin^{2}x)cosx]dx= ∫(cosx – sin

^{2}xcosx – sin^{2}xcosx + sin^{4}xcosx)dx= ∫cosx dx – 2∫sin

^{2}xcosx dx + ∫sin^{4}xcosx dxPutting sinx = t and cosxdx = dt in 2nd and 3rd integral we get:

I = ∫cos dx – 2∫t

^{2}dt + ∫t^{4}dt= sinx – 2t

^{3}/3 + t^{5}/5 + cPutting value of t:

I = = sinx – 2sin

^{3}x/3 + cos^{5}x/5 + c

### Question 4.** **∫sin^{5}xcosx dx

**Solution: **

Let I = ∫sin

^{5}xcosx dx −(i)Let sinx = t:

On differentiating with respect to x:

-cosx = dt/dx

cosx dx = -dt

Putting cosxdx = -dt and sinx = t in eq (i):

I = ∫t

^{5}dt= t

^{6}/6 + c= sin

^{6}x/6 + c

### Question 5.** **∫sin^{3}xcos^{6}x dx

**Solution: **

Let I = ∫sin

^{3}xcos^{6}x dx −(i)Let cosx = t

On differentiating both sides w.r.t′x′:

-sinx = dt/dx

sinxdx = -dt

Putting cosx = t and sinxdx = -dt in eq (i):

I = -∫sin

^{2}x t^{6}dt= -∫(1 – cos

^{2}x)t^{6}dt= -∫(1 – t

^{2})t^{6}dt= -∫(t

^{6}– t^{8})dt= -(t

^{7}/7 – t^{9}/9) + cPutting value of t:

I = -(cos

^{7}x/7 – cos^{9}x/9) + c

### Question 6.** **∫cos^{7}x dx

**Solution: **

Let I = ∫cos

^{7}x dx= ∫cos

^{6}xcosx dx= ∫(cos

^{2}x)^{3}cosx dx= ∫(1 – sin

^{2}x)^{3}cosx dx= ∫(1 – sin

^{6}x – 3sin^{2}x + 3sin^{4}x)cosx dx= ∫(cosx – sin

^{6}xcosx – 3sin^{2}xcosx + 3sin^{4}xcosx)dx −(i)Putting sinx = t and cosx dx = t in 2nd,3rd and 4th integral in (i):

I = ∫cosx dx – ∫t

^{6}dt – 3∫t^{2}dt + 3∫t^{4}dt= sinx – t

^{7}/7 - 3t^{3}/3 +3t^{5}/5 + cPutting value of t:

= sinx – sin

^{7}x/7 - 3sin^{3}x/3 +3sin^{5}x/5 + c

### Question 7.** **∫xcos^{3}x^{2}sinx^{2}dx

**Solution: **

Let I = ∫xcos

^{3}x^{2}sinx^{2}dx −(i)Let cosx

^{2 }= tOn differentiating both sides:

-2xsinx

^{2 }= dt/dx xsinx

^{2 }dx = -dt/2Putting values in (i):

= -t

^{4}/8 + cPutting value of t:

### Question 8.** **∫sin^{7}x dx

**Solution: **

Let I = ∫sin

^{7}x dxI = ∫sin

^{6}x sinx dx= ∫(sin

^{2}x)^{3}sinx dx= ∫(1 – cos

^{2}x)^{3}sinx dx= ∫(1 – cos

^{6}x – 3cos^{2}x + 3cos^{4}x)sinx dxI = ∫sinx dx – ∫cos

^{6}xsinx dx + 3∫cos^{4}xsinx dx – 3∫cos^{2}xsinx dxPutting cosx = t and sinx dx = -dt in 2nd,3rd and 4th integral:

I = ∫sinx dx – ∫t

^{6}(-dt) + 3∫t^{4}(-dt) – 3∫t^{2}(-dt)

### Question 9.** **∫sin^{3}xcos^{5}x dx

**Solution:**

Let I = ∫sin^{3}xcos^{5}x dx −(i)

Let cosx = t

On differentiating both sides: -sinx = dt/dx

sinx dx = -dt

Putting values in (i):

I = ∫sin^{2}xt^{5}(-dt)

=^{−}∫(1 – cos^{2}x)t^{5 }dt

=^{−}∫(1 – t^{2})t^{5 }dt

= ∫(t^{7 }– t^{5}) dt

= t^{8}/8 – t^{6}/6 + c

Putting value of t:

### Question 10.

**Solution:**

Let I =

Dividing and multiplying the equation by cos

^{6}x:

Let tanx = t, then:

sec^{2}x = dt/dx

sec^{2}x dx = dt

Putting values in eq (ii):

### Question 11.** **

**Solution:**

Dividing and multiplying by cos

^{8}x: Let tanx=t,then: Putting values in ii:

### Question 12.** **

**Solution:**

Dividing and multiplying by cos

^{4}x: Let tanx=t,then: sec^{2}xdx = dt Putting values in i: Putting value of t:

### Question 13.** **

**Solution:**

Let tanx=t⟹sec

^{2}x dx = dt: Putting value of t:

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