# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.25 | Set 1

### Evaluate the following integrals:

### Question 1. ∫x cosxdx

**Solution:**

Given that, I = ∫x cosxdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cosxdx – ∫(1 × ∫cosxdx)dx + c

= xsinx – ∫sinxdx + c

Hence, I = x sinx + cosx + c

### Question 2. ∫log(x + 1)dx

**Solution:**

Given that, I = ∫log(x + 1)dx

= ∫1 × log(x + 1)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log(x + 1)∫1dx – ∫(1/(x + 1) × ∫ 1dx)dx + c

= xlog(x + 1) – ∫(x/(x + 1))dx + c

= x log(x + 1) – ∫(1 – 1/(x + 1))dx + c

Hence, I = x log(x + 1) – x + log(x + 1) + c

### Question 3. ∫x^{3} logxdx

**Solution:**

Given that, I = ∫ x

^{3}logxdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = logx ∫x

^{3}dx – ∫(1/x × ∫x^{3}dx)dx + c= x

^{4}/4 logx – ∫x^{4}/4x dx+c= x

^{4}/4 logx – 1/4∫x^{3}dx + c= x

^{4}/4 logx – 1/4 ∫x^{4}/4 dx + cI = x

^{4}/4 logx – 1/16 x^{4 }+ c

### Question 4. ∫xe^{x} dx

**Solution:**

Given that I = ∫xe

^{x}dxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = xe

^{x }– ∫1.e^{x}dx= xe

^{x }– e^{x }+ cHence, I = = xe

^{x }– e^{x }+ c

### Question 5. ∫xe^{2x} dx

**Solution:**

Given that, I = ∫xe

^{2x}dxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫e

^{2x}dx – ∫(1 × ∫ e^{2x}dx) dx + c= x∫e

^{2x }dx – ∫(1 × ∫e^{2x}dx)dx + c= (xe

^{2x})/2 – ∫(e^{2x}/2)dx + c= (xe

^{2x})/2 – e^{2x}/4 + cHence, I = (x/2 – 1/4) e

^{2x }+ c

### Question 6. ∫x^{2} e^{-x} dx

**Solution:**

Given that I = ∫x

^{2}e^{-x}dxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x

^{2}∫e^{-x}dx – ∫(2x∫e^{-x}dx)dx= -x

^{2}e^{-x }– ∫(2x)(-e^{-x})dx= -x

^{2}e^{-x }+ 2∫xe^{-x}dx= -x

^{2}e^{-x }+ 2[x∫e^{-x}dx – ∫(1 × ∫ e^{-x}dx) dx]= -x

^{2}e^{-x }+ 2[x(-e^{-x}) – ∫(-e^{-x})dx]= -x

^{2}e^{-x }– 2xe^{-x }+ 2∫e^{-x}dxHence, I = -x

^{2}e^{-x }– 2xe^{-x }– 2e^{-x }+ c

### Question 7. ∫ x^{2}cosxdx

**Solution:**

Given that, I = ∫ x

^{2}cosxdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x

^{2}∫ cosxdx – ∫(2x)cosxdx)dx= x

^{2}sinx – 2∫(x)(sinx)dx= x

^{2}sinx – 2[x∫sinxdx – ∫(1 × ∫sinxdx)dx]= x

^{2}sinx – 2[x(-cosx) – ∫(-cosx)dx]= x

^{2}sinx + 2xcosx – 2∫(cosx)dxHence, I = x

^{2}sinx + 2xcosx – 2sinx + c

### Question 8. ∫x^{2}cos2xdx

**Solution:**

Given that, I = ∫x

^{2}cos2xdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x

^{2}∫cos2xdx – ∫(2x∫ cos2xdx)dx= x

^{2}(sin2x)/2 – 2∫x((sin2x)/2)dx= 1/2 x

^{2}sin2x – ∫xsin2xdx= 1/2 x

^{2}sin2x – [x∫sin2xdx – ∫ (1∫ sin2xdx)dx]= 1/2 x

^{2}sin2x – [x((-cos2x)/2) – ∫(-(cos2x)/2)dx]= 1/2 x

^{2}sin2x + x/2 cos2x – 1/2 ∫(cos2x)dxHence, I = 1/2 x

^{2}sin2x + x/2 cos2x – 1/4 sin2x + c

### Question 9. ∫xsin2xdx

**Solution:**

Given that, I =∫xsin2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫sin2xdx – ∫(1)sin2xdx)dx

= x(-(cos2x)/2) – ∫(-(cos2x)/2)dx

= -x/2 cos2x + 1/2 ∫cos2xdx

= -x/2 cos2x + 1/2(sin2x)/2 + c

Hence, I = -x/2 cos2x + 1/4 sin2x + c

### Question 10. ∫(log(logx))/x dx

**Solution:**

Given that,

I = ∫(log(logx))/x dx= ∫(1/x)(log(logx))dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = loglogx]1/x dx – ∫(1/(xlogx)∫1/x dx)dx

= logx × log(logx) – ∫(1/(xlogx) logx)dx

= logx × log(logx) – ∫1/x dx

= logx × log(logx) – logx + c

Hence, I = logx(loglogx – 1) + c

### Question 11. ∫x^{2} cosxdx

**Solution:**

Given that I = ∫x

^{2 }cosxdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x

^{2}∫ cosxdx – ∫(2x]cosxdx)dx= x

^{2}sinx – 2∫xsinxdx= x

^{2}sinx – 2[x∫sinxdx – ∫(1]sinxdx)dx]= x

^{2}sinx – 2[x(-cosx) – ∫(-cosx)dx]= x

^{2}sinx + 2xcosx – 2∫(cosx)dxHence, I = x

^{2}sinx + 2xcosx – 2sinx + c

### Question 12. ∫xcosec^{2}xdx

**Solution :**

Given that, I = ∫xcosec

^{2}xdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cosec

^{2}xdx – ∫(∫ cosec^{2}xdx)dx= -xcotx + ∫cotxdx

= -x cotx + log |sinx| + c

Hence, I = -x cotx + log |sinx| + c

### Question 13. ∫xcos^{2}xdx

**Solution:**

Given that, I = ∫xcos

^{2}xdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cos

^{2}xdx – ∫(1∫ cos^{2}xdx)dx= x∫((cos2x + 1)/2)dx – ∫(∫((1 + cos2x)/2)dx)dx

= x/2 [(sin2x)/2 + x] – 1/2∫(x + (sin2x)/2)dx

= x/4 sin2x + x

^{2}/2 – 1/2 × x^{2}/2 – 1/4 (-(cos2x)/2) + cHence, I = x/4 sin2x + x

^{2}/4 + 1/8 cos2x + c

### Question 14. ∫x^{n} logx dx

**Solution:**

Given that, I = ∫x

^{n}logxdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = logx∫x

^{n}dx – ∫(1/x ∫x^{n}dx)dx= x

^{n+1}/(n + 1) logx – ∫(1/x × x^{n+1}/(n + 1))dx= x

^{n+1}/(n + 1) logx – ∫(x^{n}/(n + 1))dxHence, I = x

^{n+1}/(n + 1) logx – 1/(n + 1)^{2}× (x^{n+1}) + c

### Question 15. ∫(logx)/x^{n} dx

**Solution:**

Given that, I = ∫(logx)/x

^{n }dx = ∫(logx)(1/x^{n})dxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = logx∫(1/x

^{n})dx – ∫((d(logx))/dx)(∫(1/x^{n})dx)dx= logx(x

^{1-n}/(1 – n)) – ∫1/x (x^{1-n}/(1 – n))dx= logx(x

^{1-n}/(1 – n)) – ∫(x^{n}/(1 – n))dx= logx(x

^{1-n}/(1 – n)) – (1/(1 – n))(x^{1-n}/(1 – n))Hence, I = logx(x

^{1-n}/(1 – n)) – (x^{1-n}/([1 – n]^{2})) + c

### Question 16. ∫x^{2} sin^{2}xdx

**Solution:**

Given that, I = ∫x

^{2}sin^{2}xdx= ∫x

^{2}((1 – cos2x)/2)dxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= ∫x

^{2}/2 dx – ∫((x^{2}cos2x)/2)dx= x

^{3}/6 – 1/2 [∫x^{2}cos2xdx]= x

^{3}/6 – 1/2 [x^{2}∫cos2xdx – ∫ (2x∫cos2xdx)dx]= x

^{3}/6 – 1/2 (x^{2}(sin2x)/2) + 1/2 × 2∫(x (sin2x)/2)dx= x

^{3}/6 – 1/4 x^{2}sin2x + 1/2 [x ∫sin2xdx – ∫(1∫sin2xdx)dx]= x

^{3}/6 – 1/4 x^{2}sin2x + 1/2 [x(-(cos2x)/2) – ∫(-(cos2x)/2)dx]= x

^{3}/6 – 1/4 x^{2}sin2x + 1/2 x(-(cos2x)/2) + 1/4 × (sin2x/2) + c= x

^{3}/6 – 1/4 x^{2}sin2x – 1/4 x(cos2x) + 1/8 × (sin2x) + cHence, I = x

^{3}/6 – 1/4 x^{2}sin2x – 1/4 x(cos2x) + 1/8 × (sin2x) + c

### Question 17.

**Solution:**

Given that, l =

Let us assume, x

^{2}= t2xdx = dt

I = ∫t × e

^{t}dtUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= t∫e

^{t}dt – ∫(1 × ∫e^{t}dt)dt= te

^{t }– ∫e^{t}dt= te

^{t }– e^{t }+ c= e

^{t-1 }+ cHence, I = (x

^{2}– 1) + c

### Question 18. ∫x^{3} cosx^{2} dx

**Solution:**

Given that, I = ∫x

^{3}cosx^{2}dxLet us assume x

^{2 }= t2xdx = dt

I = 1/2 ∫tcostdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/2[t∫costdt – ∫(1 × ∫costdt)dt]

= 1/2 [t × sint – ∫sintdt]

= 1/2[tsint + cost] + c

Hence, I = 1/2 [x² sinx

^{2 }+ cosx^{2}] + c

### Question 19. ∫xsinxcosxdx

**Solution:**

Given that, I = ∫xsinxcosxdx

= ∫x/2(2sinxcosx)dx

= 1/2 ∫xsin2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/2 [x∫sin2xdx – ∫(1 × ∫sin2xdx)dx]

= 1/2 [x((-cos2x)/2) – ∫((-cos2x)/2)dx]

= -1/4 xcos2x + 1/4 ∫cos2xdx

Hence, I = -1/4 xcos2x + 1/8 sin2x + c

### Question 20. ∫sinx(logcosx)dx

**Solution:**

Given that, I = ∫sinx(logcosx)dx

Let us considered, cosx = t

-sinxdx = dt

I = -∫ logtdt

= -∫1 × logtdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= -[logt∫dt – ∫(1/t × ∫dt)dt]

= -[tlogt – ∫1/t × tdt]

= -[tlogt-∫ dt]

= -[tlogt – t + c

_{1}]= t(1 – logt) + c

Hence, I = cosx(1 – logcosx) + c

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