Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.6

Integrate the functions in Exercises 1 to 22.

Question 1: x sinx.

Solution:

Let f(X) = ∫ x sinx dx

Taking x as first function and sin x as second function and integrating by parts, we obtain

f(x) = x ∫sin x dx – {(d(x)/dx) ∫sin x dx} dx

⇒ f(x) = x(-cosx)-∫1. (-cosx)dx

⇒ f(x) = x cosx + sinx + C

Question 2: x sin3x

Solution:

Let f(x) = ∫x sin3x dx

Taking x as the first function and sin 3x as the second function and integrating by parts, we obtain

f(x) = x ∫sin3x dx – {(d(x)/dx) ∫sin3x dx} dx

⇒ f(x) = x (-cos3x ∕ 3) – ∫1. (cos3x ∕ 3) dx

⇒ f(x) = -x cos3x∕ 3 + 1∕3 ∫cos3x dx

⇒ f(x) = -xcos3x∕3 + 1∕9 sin3x + C

Question 3: x² e^x

Solution:

Let f(x) = ∫x² e^x dx

Taking x² as first function and e^x as second function and integrating by parts, we obtain

f(x) = x² ∫e^x dx – {(d(x²)/dx) ∫e^x dx} dx

⇒ f(x) = x² e^x– ∫2x.e^x dx

⇒ f(x) = x²e^x – 2 ∫x.e^x dx

Again, integrating by parts, we obtain

f(x) = x²e^x -2[x.∫e^xdx-∫{(d(x)/dx). ∫e^xdx}dx]

⇒ f(x) = x²ex -2[xe^x – ∫e^xdx]

⇒ f(x) = x²e^x-2[xe^x – e^x]

⇒ f(x) = x²e^x -2xe^x + 2e^x + C

⇒ f(x) = e^x (x²-2x+2) + C

Question 4: x logx

Solution:

Let f(x) = ∫x log x dx

Taking log x as first function and x as second function and integrating by parts, we obtain

f(x) = log x ∫x dx – {(d(log x)/dx) ∫x dx} dx

⇒ f(x) = log x. (x² ∕ 2) – ∫1∕x. (x² ∕2) dx

⇒ f(x) = x²logx∕2 – ∫x∕2 dx

⇒ f(x) = x²logx∕2 – x²∕4 + C

Question 5: x log2x

Solution:

Let f(x) = ∫x log 2x dx

Taking log 2x as first function and x as second function and integrating by parts, we obtain

f(x) = log 2x ∫x dx – {(d (log 2x)∕dx) ∫x dx} dx

⇒ f(x) = log 2x. (x² ∕ 2) – ∫2∕2x. (x² ∕2) dx

⇒ f(x) = x²log2x∕2 – ∫x∕2 dx

⇒ f(x) = x²log2x∕2 – x2∕4 + C

Question 6: x² logx

Solution:

Let f(x) = ∫x² log x dx

Taking log x as first function and x² as second function and integrating by parts, we obtain

f(x) = log x ∫x² dx – {(d (log x) ∕dx) ∫x² dx} dx

⇒ f(x) = log x. (x³ ∕ 3) – ∫1∕x. (x³ ∕3) dx

⇒ f(x) = x³logx∕3 – ∫x³∕ 3 dx

⇒ f(x) = x³logx∕3 – x³ ∕ 9 + C

Question 7: x sin^-1x.

Solution:

Let f(x) = ∫x sin^-1x dx

Taking sin^-1x as first function and x as second function and integrating by parts, we obtain

f(x) = sin^-1x ∫ x dx – ∫ {(d(sin^-1x∕dx) ∫x dx} dx

⇒ f(x) = sin^-1x (x²/2) – ∫ 1∕ √(1-x² ).x²∕2 dx

⇒ f(x) = x²sin^-1x ∕2 + 1∕2 ∫-x ∕ √(1-x² ) dx

⇒ f(x) = x² sin^-1x∕2 + 1∕2 ∫ {1-x² ∕ √(1-x² )- 1∕√(1-x²)} dx

⇒ f(x) = x² sin^-1x∕2 + 1∕2 ∫ {√1-x² – 1∕√(1-x²)} dx

⇒ f(x) = x² sin-1x∕2 + 1∕2 {∫ √1-x² dx – 1 ∕√(1-x²) dx}

⇒ f(x) = x²sin^-1x∕2 + 1∕2{x∕2. √1-x² + 1∕2sin^-1x – sin^-1x} +C

⇒ f(x) = x²sin^-1x∕2 + x∕4. √1-x² + 1∕4sin^-1x – 1∕2sin^-1x +C

⇒ f(x) = 1∕4(2x²-1) sin^-1x + x∕4. √1-x² + C

Question 8: x tan^-1x

Solution:

Let f(x) = ∫ x tan^-1x dx

Taking tan^-1x as first function and x as second function and integrating by parts, we obtain

f(x) = tan^-1x ∫ x dx – ∫ {(d(tan^-1x∕dx) ∫x dx} dx

⇒ f(x) = tan^-1x (x²∕2) – ∫ 1∕ (1+x). x²∕2 dx

⇒ f(x) = x² tan^-1x ∕ 2 – 1∕2∫ 1∕ (1+x² ) dx

⇒ f(x) = x² tan^-1x ∕ 2 – 1∕2∫{(x²+1) ∕ (1+x²) – 1∕ (1+x²)} dx

⇒ f(x) = x² tan^-1x ∕ 2 – 1∕2∫ (1- 1∕ (1+x²) dx

⇒ f(x) = x² tan^-1x ∕ 2 – 1∕2 (x-tan^-1 x) + C

⇒ f(x) = x² tan^-1x ∕ 2 – x∕2 + 1∕2 tan^-1x + C

Question 9: x cos^-1x

Solution:

Let f(x) = ∫x cos^-1x dx

Taking cos^-1x as first function and x as second function and integrating by parts, we obtain

f(x) = cos^-1x ∫ x dx – ∫ {(d(cos^-1x)∕dx.∫x dx} dx

⇒ f(x) = cos^-1x (x²/2) – ∫ -1∕ √1-x² .x²∕2 dx

⇒ f(x) = x² cos^-1x ∕2 – 1∕2 ∫1-x²-1 ∕ √(1-x² ) dx

⇒ f(x) = x² cos^-1x∕2 – 1∕2 ∫ {√(1-x² ) +( -1∕√1-x²)} dx

⇒ f(x) = x² cos^-1x∕2 – 1∕2 ∫ √1-x² dx – 1∕2 ∫(-1∕√1-x²) dx

⇒ f(x) = x² cos^-1x∕2 – 1∕2 I₁ – 1∕2 cos^-1x ————-(1)

Where I₁ = ∫√1-x² dx

I₁ = x√1-x² – ∫d(√1-x²)∕dx ∫x dx

⇒ I₁ = x√1-x² – ∫d(-2x∕2√1-x² .x dx

⇒ I₁ = x√1-x² – ∫-x²∕√1-x² dx

⇒ I₁ = x√1-x² – ∫1-x²-1 ∕ √(1-x²) dx

⇒ I₁ = x√1-x² – { ∫ √1-x² dx + ∫(-dx∕√1-x²}

⇒ I₁ = x√1-x²– {I₁ + cos^-1x}

⇒ 2I₁ = x√1-x² – cos^-1x

⇒ I₁ = x∕2 .√1-x² -1∕2 cos^-1x

Substituting in (1), we obtain

f(x) = x²cos^-1x∕2 – 1∕2 (x∕2 .√1-x² -1∕2 cos^-1x) – 1∕2 cos^-1x

⇒ f(x) = (2x-1)∕4 cos^-1x – x∕4 √1-x² + C

Question 10: (sin^-1x)²

Solution:

Let f(x) = ∫(sin^-1x)² dx

Taking (sin^-1x)²as first function and 1 as second function and integrating by parts, we obtain

f(x) = (sin^-1x)²∫ 1dx – ∫ {(d(sin^-1x)²∕dx. ∫1 dx} dx

⇒ f(x) = (sin^-1x). x – ∫ 2.sin^-1x∕√1-x² . x dx

⇒ f(x) = x(sin^-1x) ² + ∫sin^-1x.(-2x∕ √1-x²⁾dx

⇒ f(x) = x(sin^-1x) ²+[sin^-1x∫-2x∕√1-x² dx – ∫ {❴d(sin^-1x)∕dx❵∫-2x ∕√1-x² dx}dx]

⇒ f(x) =x(sin^-1x) ²+[ sin^-1x .2√1-x² – ∫1∕√1-x² .2√1-x² dx ]

⇒ f(x) =x(sin^-1x)²+ 2√1-x² sin^-1x – ∫2dx

⇒ f(x) =x(sin^-1x)²+ 2√1-x² .sin^-1x -2x + C

Question 11: (x cos^-1x) / √1-x²

Solution:

Let f(x) = ∫(x cos^-1x) / √1-x² dx

we are multiplying -1/2 in numerator and dementor then.

f(x) = -1∕2 ∫(-2x cos^-1x) ∕√1-x² dx

Taking (cos^-1x) as first function and {-2x∕√1-x² ❵ as second function and integrating by parts, we obtain

f(x) = -1∕2 [ cos^-1x∫-2x∕√1-x dx – ∫ {❲d(cos^-1x)∕dx❳∫-2x√1-x² dx}dx]

⇒ f(x) = -1∕2 [cos^-1x.2√1-x² – ∫-1∕√1-x² 2√1-x² dx]

⇒ f(x) = -1∕2 [2√1-x² cos^-1x+ ∫2 dx]

⇒ f(x) = -1∕2 [2√1-x² cos^-1x+ 2x] + C

⇒ f(x) = – [√1-x² cos^-1x+ x] + C

Question 12: x sec²x

Solution:

Let f(x) = ∫x sec²x dx

Taking x as first function and sec²x as second function and integrating by parts, we obtain

f(x) = x ∫sec²x dx – {(d (x) ∕dx) ∫sec²x dx} dx

⇒ f(x) = xtanx – ∫1.tanx dx

⇒ f(x) = x tanx + log|cosx| + C

Question 13: tan^-1x

Solution:

Let f(x) = ∫tan^-1x dx

Taking tan^-1x as first function and 1 as second function and integrating by parts, we obtain

f(x) = tan^-1x∫1dx – ∫ {[d❲tan^-1x❳∕dx] ∫1. dx}dx

⇒ f(x) = tan^-1x .x – ∫1∕1+x² .x dx

⇒ f(x) = x tan^-1x – 1∕2 ∫2x∕1+x² dx

⇒ f(x) = x tan^-1x – 1∕2 log|1+x²| +C

⇒ f(x) = x tan^-1x -1∕2 log(1+x²) + C

Question 14: x (logx)²

Solution:

Let f(x) = ∫ x (logx)² dx

Taking (logx)² as first function and x as second function and integrating by parts, we obtain

f(x) = (logx)² ∫x dx – ∫[{{d(logx)∕dx}²}∫xdx]dx

⇒ f(x) = x∕2 (logx)² – [∫2logx 1∕x . x²∕2 dx]

⇒ f(x) = x²∕2 (logx)² – ∫x logx dx

Again, integration by parts, we obtain.

f(x) = x²∕2 ❲logx❳² – [logx ∫x dx – ∫{❴d(logx)∕dx}∫xdx❵dx]

⇒ f(x) = x²∕2 ❲logx❳² – [x²∕2 – logx – ∫1∕x .x²∕2 dx]

⇒ f(x) = x²∕2 ❲logx❳² – x²∕2 .logx + 1∕2∫xdx

⇒ f(x) = x²∕2 ❲logx❳² – x²∕2 .logx + x²∕4 + C

Question 15: (x²+1) logx

Solution:

Let f(x) = ∫ (x²+1) logx dx

⇒ f(x) = ∫ x² logx dx + ∫ logx dx

Let f(x) = I₁ + I₂ ……………………….. (1)

where I₁ = ∫ x² logx dx and I₂ = ∫ logx dx

I₁ = ∫ x² logx dx

Taking (logx) as first function and x² as second function and integrating by parts, we obtain

⇒ I₁ = logx – ∫ x²dx – ∫{❴d(logx)∕dx❵∫x²dx} dx

⇒ I₁ = logx .x³∕3 – ∫1∕x . x³∕3 dx

⇒ I₁ = x³∕3 logx – 1∕3(∫x dx)

⇒ I₁ = x³∕3 logx – x³∕9 + C₁ …………………… (2)

I₂ = ∫ logx dx

Taking log x as first function and 1 as second function and integrating by parts, we obtain

f(x) = log x ∫1 dx – {(d(log x)/dx) ∫1 dx} dx

⇒ f(x)= log x.x – ∫1∕x. x dx

⇒ f(x) = x. logx∕2 – ∫1 dx

⇒ f(x) = x. logx∕2 – x + C₂ ………………….(3)

Using equation (2) and (3) in (1), we obtain

f(x) = x³∕3 logx – x³∕9 + C1 + x. logx∕2 – x + C2

⇒ f(x)= x³∕3 logx – x³∕9 + x. logx∕2 – x + C ₁+ C2

⇒ f(x)= (x³∕3 + x) logx – x³∕9 – x + C

Question 16: e^x(sinx + cosx)

Solution:

Let f(x) = ∫ e(sinx+cosx) dx

Let g(x) = sinx

g^‘(x) = cosx

f(x) = ∫ e^x{g(x) + g^‘(x) } dx

It is known that, ∫ e^x{ g(x) + g'(x) } dx = e^x g(x) + C

So, f(x) = e^x sinx + C

Question 17: x e^x ∕(1+x)²

Solution:

Let f(x) = ∫ x e^x ∕(1+x)² dx

⇒ f(x)= ∫ e^x{x ∕(1+x)²} dx

⇒ f(x)= ∫ e^x{(1+x-1) ∕ (1+x)²} dx

⇒ f(x)= ∫ e {1∕ (1+x) – 1∕ (1+x)²} dx

Let f(x) = 1∕ (1+x)

f^‘(x) = -1∕ (1+x)²

⇒ f(x) = ∫ x e^x ∕(1+x)² dx = ∫ e^x{f(x) + f'(x) }dx

It is known that ∫ e^x{f(x) + f'(x) }dx = e^x f(x) + C

So, ∫ x e^x ∕(1+x)² dx = e^x ∕ (1+x) C

Question 18: e^x{(1+sinx) ∕ (1+cosx)}

Solution:

Let I = e^x{(1+sinx)∕(1+cosx)}

⇒ I =e^x{(sin²x∕2+cos²x∕2+2 sinx∕2 cosx∕2) ∕ (2cos²x∕2)}

⇒ I = e^x{(sinx∕2+cosx∕2)² ∕ (2cos²x∕2)}

⇒ I = 1∕2 e^x{(sinx∕2+cosx∕2) ∕ (cosx∕2)}²

⇒ I = 1∕2 e^x {tanx∕2 + 1}²

⇒ I = 1∕2 e^x{1 + tan²x∕2 + 2tanx∕2}

⇒ I = 1∕2 e^x{secx∕2 + 2tanx∕2}

⇒ I = e^x(1+sinx)dx ∕ (1+cosx) = e^x{1∕2 sec²x∕2 + tanx∕2} —————– (1)

Let tanx∕2 = f(x) or f'(x) = 1∕2 sec²x∕2

It is known that ∫ e^x{f(x) + f'(x) } dx = e^x f(x) + C

from equation (1), we obtain,

∫e^x{(1+sinx)∕(1+cosx)}dx = e^x tanx∕2 + C

Question 19: e^x{1∕x – 1∕x²}

Solution:

Let f(x) = ∫e^x{1∕x – 1∕x²} dx

Also Let 1∕x = f(x) or f'(x) = – 1∕x²

It is known that ∫ e^x{f(x) + f'(x) } dx = e^x f(x) + C

So, f(x) = e^x ∕ x + C

Question 20: (x-3) e^x ∕ (x-1)³

Solution:

Let f(x) = ∫ e^x{ (x-3) ∕ (x-1)³}dx

⇒ f(x) = ∫ e^x{ (x-1-2) ∕ (x-1)³}dx

⇒ f(x) = ∫ e^x{ 1∕ (x-1)² – 2 ∕ (x-1)³}dx

⇒ f(x) = 1 ∕ (x-1)² or f'(x) = -2 ∕ (x-1)³

It is known that ∫ e^x{f(x) + f'(x) } dx = e^x f(x) + C

So, ∫ e^x{ (x-3) ∕ (x-1)³}dx = e^x ∕ {x-1}² + C

Question 21: e^2x sinx

Solution:

Let f(x) = ∫ e^2x sinx dx ————– (1)

Integrating by parts, we obtain

f(x) = sinx ∫ e^2x dx – ∫ { ❴ d(sinx)∕dx❵ ∫ e^2x dx} dx

⇒ f(x) = sinx . e^2x∕2 – ∫ cosx e^2x∕2 dx

⇒ f(x) = 1∕2 e^2x sinx – 1∕2 ∫ e^2x cosx dx

Again, Integrating by parts, we obtain

f(x) = 1∕2 e^2x sinx – 1∕2 [cosx ∫ e^2x dx – ∫ {❴d∕dx cosx❵∫e^2x dx} dx]

⇒ f(x) = 1∕2 e^2x sinx – 1∕2 [cosx e^2x ∕ 2 – ∫ ❴-sinx❵ e^2x ∕2 dx

⇒ f(x) = 1∕2 e^2x sinx – 1∕2 [(cosx e^2x ) ∕ 2 + ∫ sinx e^2x ∕2 dx

⇒ f(x) = 1∕2 e^2x sinx – (e^2x cosx) ∕ 4 – 1∕4f(x) ———-from (1)

⇒ f(x) + 1∕4f(x) = 1∕2 e^2x sinx – (e^2x cosx) ∕ 4

⇒ 5/4 f(x) = (e^2x sinx)1∕2 – (e2x cosx) ∕ 4

⇒ f(x) = e^2x ∕5 [ 2 sinx – cosx] + C

Question 22: sin^-1(2x∕(1+x²)

Solution:

sin^-1(2x∕(1+x²)

Let x = tanθ or dx = sec²θ dθ

So, sin^-1(2x∕(1+x²) = sin^-1(2 tanθ∕(1+tan²θ) = sin^-1(sin2θ) = 2θ

Integrating by parts, we obtain

2[θ.∫sec²θ dθ – ∫{❴dθ∕ dθ❵sec²θ dθ} dθ]

= 2[θ.tanθ – ∫tanθ dθ]

= 2[θ.tanθ – log|cosθ|] + C

= 2[x.tan^-1x – log|1√(1+x²|] + C

= 2xtan^-1x + 2 log(1+x²)^1∕2 + C

= 2x tan^-1x + 2 [-1∕2 log(1+x²) ] + C

= 2x tan^-1x – log(1+x²) + C

Choose the correct answer in Exercises 23 and 24.

Question 23: ∫ x²e^{x^3} dx equals

(A) 1∕3 .e^{X^3} + C

(B) 1∕3 . e^{X^3} + C

(C) 1∕2 .e^{X^3} + C

(D) 1∕3 . e^{X^3} + C

Correct answer is A.

Question 24 :∫ e^x secx(1+tanx) dx equals.

(A) e^x cosx + C

(B) e^x secx + C

(C) e^x sinx + C

(D) e^x tanx + C

Correct answer is (B) e^x secx + C.