Indefinite Integrals Class 12 OP Malhotra Exe-14A ISC Maths Solutions Ch-14 Solutions. In this article you would learn about methods of integration and method of substitution. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Indefinite Integrals Class 12 OP Malhotra Exe-14A ISC Maths Solutions Ch-14

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-14	Indefinite Integrals
Writer	OP Malhotra
Exe-14(a)	Methods of Substitution for Integration

Methods of  Integration by Substitution

Indefinite Integrals Class 12 OP Malhotra Exe-14A ISC Maths Solutions Ch-14 Solutions

Que-1: ∫6x-8/3x²-8x+5 dx

Sol: Let I = ∫6x−8/3x²−8x+5 dx
Put 3x² – 8x + 5 = t ⇒(6x – 8) dx = dt
= ∫dt/t = log |t| + c
= log|3×2 – 8x + 5| + c

Que-2: ∫dx/(3-5x) dx

Sol: ∫dx / 3−5x; Put 3 – 5x = t
⇒ -5dx = dt
= ∫dt/−5t = −1/5log|t| + c
= –1/5log|3 – 5x| + c

Que-3: ∫√(1+x) dx

Sol: ∫√(1+x) dx = ∫(1+x)^1/2 dx
= (1+x)^1/2+1/(1/2+1) + c
=2/3 (1+x)^3/2 + c

Que-4: ∫cosec²x/(1+cotx) dx

Sol:

Que-5: ∫(cosx−sinx)/(cosx+sinx) dx

Sol: Let I = ∫(cosx−sinx)/(cosx+sinx) dx
Put cos x + sin x = t
⇒ (c – sin x + cos x) dx = dt
∴ I = ∫dt/t = log | t | + c
= log |cos x + sin x| + c

Que-6: ∫secx log(secx+tanx) dx

Sol: Let I = ∫secx log(secx+tanx) dx
Put log (sec x + tan x) = t;
Differentate both sides w.r.t. x, we have
⇒ 1/secx+tanx (sec x tan x + sec² x)dx = dt
⇒ secx(secx+tanx)/(secx+tanx) dx = dt
⇒ sec x dx = dt
∴ I = ∫t.dt = t²/2 + c
= 1/2 [log(sec x + tan x]² + c

Que-7: ∫cos√x / √x dx

Sol: Put √x = t ⇒ 1/2√x dt = dt
∴ I = ∫cos√x/√x dx = ∫cost(2di)
= 2 sin t + c
= 2 sin√x + c

Que-8: ∫xe^{x^(2)} dx

Sol: Let I = ∫xe^{x^(2)} dx;
Put x² = t ⇒ 2xdx = dt
= ∫e^t dt/2 = 1/2 e^t+c
= 1/2 e^{x^(2)}+c

Que-9: ∫e^(m tan^-1x) / 1+x² dx

Sol: Let I = ∫e^(m tan^-1x) / 1+x² dx
Put tan^-1x = t ⇒ 1/1+x² dx = dt
∴ I = ∫e^mt dt = e^mt/m + c
= e^(m tan^-1x) /m + c

Que-10: ∫sec²(logx) / x dx

Sol: Let I = ∫sec²(logx)/x dx
Put log x = t ⇒ 1/x dx = dt
∴ I = ∫sec²t dt = tan t + c
= tan (log x) + c

Que-11: ∫sin²x cos x dx

Sol: Let I = ∫sin²x cos x dx
Put sin x = t ⇒ cos x dx = dt
= ∫t²dt = t³/3 + c
= sin³x/3 + c

Que-12: ∫(sin^-1x)³/√(1−x²) dx

Sol: Let I = ∫(sin^-1x)³/√(1−x²) dx
= ∫(sin^-1x)³ . 1/√(1−x²) dx
= (sin^-1x)⁴ /4 +c

Que-13: ∫ sec⁴x tan x dx

Sol: Let I = ∫sec⁴x tan x dx
= ∫sec³x (sec x tan x dx)
Put sec x = t ⇒ sec x tan x dx = dt
= ∫t³dt = sec⁴x/4 +c
= sec⁴x/4 + c

Que-14: ∫x³/(x²+1)³dx

Sol:

Que-15: ∫1+tanx/x+logsecx dx

Sol: Let I = ∫1+tanx/x+logsecx dx
Put x + log sec x = t; on differentiating
=∫t³dt= t⁴/4 + c
= sec⁴x/4+c

Que-16: ∫sec⁴x dx

Sol: ∫sec⁴x dx = ∫sec²(1 + tan² x) dx
= ∫(1+t²)dt = t+t³/3 +c
= tan x + tan³x/3 +c

Que-17: ∫sin(logx)/x dx

Sol: Let I = ∫sin(logx)/x dx;
Put log x = t ⇒ 1/x dx = dt
= ∫(sint dt) = -cos t + c
= – cos(log x) + c

Que-18: ∫cosx−sinx/(1+sin2x) dx

Sol:

 

Que-19: ∫(1+sinx/1+cosx) dx

Sol:

Que-20: ∫(sinθ+cosθ)/√sin2θ dθ

Sol: Let I = ∫(sinθ+cosθ)/√sin2θ dθ
Put sin θ – cos θ = t …(1)
⇒ (cos θ + sin θ) dθ = dt
On squaring eqn (1) both sides; we have (sin θ – cos θ)² = t²
⇒ sin²θ + cos²θ – 2 sin θ cos θ = t²
⇒ 1 – sin 2θ = t²
⇒ sin 2θ = 1 – t²

Que-21: ∫cos x /(cos x/2 + sin x/2) dx

Sol:

Que-22: ∫sinx cosx /sin⁴x+cos⁴x dx

Sol:

Que-23: ∫tanx/secx+tanx dx

Sol: Let I = ∫tanx/secx+tanx dx
= ∫tanx(secx−tanx)/sec²x−tan²x dx
= ∫tanx sec x dx – ∫tan²x dx
=∫tanxsecxdx−∫(sec²x−1)dx
= sex x – tan x + x + c

Que-24: ∫tan⁴x dx

Sol:

Que-25: ∫√1+√x dx

Sol:

Que-26: ∫x/√x+5 dx

Sol:

Que-27: ∫(x+1)(x+logx)²/x dx

Sol:

Que-28: ∫x²e^{x^(3)} cos (e^{x^(3)}) dx

Sol: Let I = ∫x²e^{x^(3)} cos (e^{x^(3)}) dx
Put e^{x^(3)} = t ⇒ e^{x^(3)} × 3 x² dx=dt
= ∫cos t dt/3 =1/3 sint+c
= 1/3 sin(e^{x^(3)})+c

Que-29: ∫2 sinθ.cosθ /sin⁴θ+cos⁴θ dθ

Sol: Let I =∫2 sinθ.cosθ /sin⁴θ+cos⁴θ dθ
Divide numerator and denominator by cos4 θ; we have

Que-30: ∫1/(9+16 cos²x) dx

Sol: Let I =∫1/(9+16 cos²x) dx
Divide numerator and denominator by cos² x; we have

Que-31: ∫sinx/a+bcosx dx

Sol:

Que-32: ∫e^x-1/e^x+1 dx

Sol:

Que-33: ∫e^2x/e^2x-2 dx

Sol:

Que-34: ∫e^2x/e^x-1 dx

Sol: Let I = ∫e^2x/e^x-1 dx
Put e^x– 1 = t ⇒ e^x = t + 1
⇒ e^x dx = dt
= ∫(t+1)/t dt = ∫(1+1/t)dt
= t + log | t | + e
= e^x + log |e^x-1| + c’ [where c’ = c – 1]

Que-35: ∫e^2x–e^-2x/e^2x+e^-2x dx

Sol: Let I = ∫e^2x-e^-2x/e^2x+e^-2x dx
Put e^2x+e^-2x = t ⇒ (2e^2x – 2e^-2x) dx = dt
⇒ (e^2x-e^-2x) dx = dt/2
= ∫dt/2t = 1/2 log |t| + c
= 1/2 log|e^2x+e^-2x| + c

Que-36: ∫cot³x cosec⁴x dx

Sol: Let I = ∫cot³x cosec⁴x dx
= ∫cot³x cosec²x (1+cot²x) dx
Put cot x = t ⇒ -cosec²x dx = dt
= −∫t³(1+t²)dt
= – t⁴/4 – t⁴/4 + c
= – 1/4 cot⁴x – 1/6 cot⁶x + c

Que-37: ∫√e^x-4 dx

Sol: Let I = ∫√e^x-4 dx
Put √e^x-4=t; on squaring both sides
e^x-4 = t² ⇒ e^x = t² + 4 ⇒ e^x dx = 2t dt

Que-38: ∫x/e^{x^(2)} dx

Sol:

Que-39: Evaluate:

Sol:

Que-40: ∫((x³-1)^1/3 x⁵ dx

Sol:

Que-41: ∫sin³x cos⁴x dx

Sol: Let I = ∫sin³x cos⁴x dx Let I = ∫sinx(1−cos²x)cos⁴x dx
Put cos x = t ⇒ – sin x dx = dt = – −∫(1−t²)t⁴ dt
= – t⁵/5 + t⁷/7 + C = – cos⁵x/5 + cos⁷x/7 + C

–: End of Indefinite Integrals Class 12 OP Malhotra Exe-14A ISC Math Ch-14 Solution :–

Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions

Please share with your friends
Thanks