Indefinite Integrals Exe-15G Class 12 OP Malhotra ISC Maths Solutions Ch-15. In this article you would learn about the integrals of the form ∫acosx+bsinx/ccosx+dsinx dx . 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-15G ISC Maths Solutions Ch-15

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-15	Indefinite Integrals
Writer	OP Malhotra
Exe-15(g)	integrals of the form ∫acosx+bsinx/ccosx+dsinx dx

Integrals of the form ∫acosx+bsinx/ccosx+dsinx dx

 Indefinite Integrals Class 12 OP Malhotra Exe-15G Solutions

Que-1: (i) ∫dx/4+5cosx
(ii) ∫dx/12+12cosx
(iii) ∫dθ/4cosθ−1
(iv) ∫dθ/1+2cosθ

Sol: (i) ∫dx/4+5cosx
Put tan x² = t ⇒ sec²x/2 1/2 dx = dt
⇒ dx 2dt/1+t²

(ii) ∫dx/12+12cosx
Let I = ∫dx/12+12cosx = ∫dx/12×2cos²x/2
= 1/24∫sec²x/2dx = 1/24(tanx/2)/12 + C
= 1/12 tanx/2 + C

(iii) ∫dθ/4cosθ−1
Let I = ∫dθ/4cosθ−1
Put tan θ/2 = t ⇒ sec²θ/2 1/2 dθ = dt
⇒ dθ = 2dt/1+t²
∴cos θ = 1−tan²θ/2 / 1+tan²θ/2 = 1−t² / 1+t²

(iv) ∫dθ/1+2cosθ
Put tan θ/2 = t ⇒ sec²θ/2 . 1/2 dθ = dt
⇒ dθ = 2dt/1+t²
& cos θ = 1−tan²θ/2 / 1+tan²θ/2 = 1−t² / 1+t²

Que-2: (i) ∫dx/4−3sinx
(ii) ∫dx/4+5sinx
(iii) ∫dθ/1−2sinθ
(iv) ∫dx/1−3sinx

Sol: (i) Let I = ∫dx/4−3sinx
Put tan x/2 = t ⇒ sec²x/2 . 1/2 dx = dt
⇒ dx = 2dt/1+t²
& tan x = 2tanx/2 / 1+tan²x/2 = 2dt/1+t²

(ii) Let I = ∫dx/4+5sinx
Put tanx/2 = t ⇒ sec²x/2 . 1/2dx = dt
⇒ dx = 2t/1+t²
& sin x = 2tanx/2 1+tan²x/2 = 2dt/1+t²

(iii) Let I = ∫dθ/1−2sinθ
Put tan θ/2 = t ⇒ sec²θ/2 . 1/2 dθ = dt
⇒ dθ = 2dt/1+t²
& sin θ = 2tanθ/2 / 1+tan²θ/2 = 2t/1+t²

(iv) Let I = ∫dx/1−3sinx
Put tanx/2 = t ⇒ sec²x/2 . 1/2dx = dt
⇒ dx = 2dt/1+t²
& sin x = 2tanx/2 / 1+tan²x/2 = 2dt/1+t²
∴ I = ∫2/1+t²  / 1−3(2t/1+t²) dt = ∫2dt/t²+1−6t
= 2∫dt/t²−6t+9−8
= 2∫dt/(t−3)²−(√8)²
= 2 × 1/2√8 log∣t−3−√8 / t−3+√8∣ + C
= 1/2√2 log ∣tanx/2−3−2√2 / tanx/2−3+2√2∣ + C

Que-3: (i) ∫dx/2sinx−3cosx
(ii) ∫dx/3cosx+4sinx
(iii) ∫dx/sinx−cosx 

Sol: (i) Let I = ∫dx/2sinx−3cosx
Put tan x/2 = t ⇒ sec²x/2 . 1/2dx = dt
⇒ dx = 2dt/1+t²
sin x = 2tanx/2 / 1+tan²x/2 = 2t/1+t²
& cos x = 1−tan²x/2 / 1+tan²x/2 = 1−t²/1+t²
∴ I = ∫2dt/1+t² / 2(2t/1+t²)−3(1−t²/1+t³)
= ∫2dt/4t−3+3t² = 2/3
= 2/3∫dt/t²+4t/3+4/9−4/9−1
= 2/3∫dt(t−2/3)²−(√13/3)²

(ii) Let I = ∫dx/3cosx+4sinx
Put tanx/2 = t ⇒ sec²x/2 . 1/2dx = dt
⇒ dx = 2dt/1+t²

(iii) Let I = ∫dx/sinx−cosx
Put tanx/2 = t ⇒ sec²x/2 . 1/2dx = dt
⇒ dx = 2dt/1+t²
∴ sin x = 2tanx/2 / 1+tan²x/2 = 2t/1+t²
& cos x = 1−tan²x/2 / 1+tan²x/2 = 1−t²/1+t²
∴ I = ∫2dt/1+t² / 2t/1+t² −1−t²/1+t²
= ∫2dt/2t−1+t² = ∫2dt/t²+2t+1−2
= ∫2dt/(t+1)²−(√2)²
= 2∫dt/u²−(√2)²
[Put t + 1 = u ⇒ dt = du]
= 2 × 1/2√2 log ∣u−√2/u+√2∣+ C
= 1/√2 log∣tanx/2+1−√2 / tanx/2+1+√2∣ + C

Que-4: (i) ∫dx/3+2sinx+cosx
(ii) ∫dx/1−sinx+cosx
(iii) ∫dx/1+sinx+cosx

Sol: (i) Let I = ∫dx/3+2sinx+cosx
Put tanx2 = t ⇒ sec²x/2 1/2dx = dt
⇒ dx = 2dt/1+t²
sin x = 2tanx/2 / 1+tan²x/2 = 2t/1+t²
& cos x = 1−tan²x/2 / 1−tan²x/2 = 1−t² / 1+t²
∴ I = ∫2dt/1+t² / 3+2(2t/1+t²)+1−t² / 1+t²
= ∫2dt/3(1+t²)+4t+1−t² = ∫2dt/2t²+4t+4
= ∫2dt/2(t²+2t+2) = ∫2dt/(t+1)²+1
= tan^-1(t + 1) + C
= tan^-1(1 + tanx/2) + C

(ii) Let I = ∫dx/1−sinx+cosx
Put tanx/2 = t ⇒ sec²x/2 1/2dx = dt
⇒ dx = 2dt/1+t²
sin x = 2tanx/2 / 1+tan²x/2 = 2t/1+t²
& cos x = 1−tan²x/2 / 1+tan²x/2 = 1−t²/1+t²
∴ I = ∫2dt/1+t² / 1−(2t/1+t²)+(1−t²/1+t²)
= ∫2dt/1+t²−2t+1−t²
= ∫dt/1−t = log|1−t|/−1 + C
= – log | 1 – tanx/2 | + C

(iii) Let I = ∫dx/1+sinx+cosx
Put tanx/2 = t ⇒ sec²x/2 1/2dx = dt
⇒ dx = 2dt/1+t²
sin x = 2tanx/2 / 1+tan²x/2 = 2t/1+t²
& cos x = 1−tan²x/2 / 1+tan²x/2 = 1−t²/1+t²
∴ I = ∫2dt1+t21+2t1+t2+1−t21+t2
= ∫2dt/1+t²+2t+1−t² = ∫dt/1+t
= log |t + 1| + C
= log | 1 + tanx/2 | + C

Que-5: (i) ∫2sinθ+cosθ/7sinθ−5cosθ dθ
(ii) ∫cosx+3sinx−5/3(1−sinx)−cosx dx

Sol: (i) Let I = ∫2sinθ+cosθ7sinθ−5cosθ dθ
Let numerator = l(deno) + m d/dθ(deno)
⇒ 2 sinθ + cosθ = l(7sinθ – 5cosθ) + m d/dθ(7 sinθ – 5cosθ)
⇒ 2 sinθ + cosθ = l(7sinθ – 5cosθ) + m(7cosθ + 5 sinθ) …(i)
Comparing the coeff. of sinθ & cosθ on both sides; we have
2 = 7l + 5m ⇒ 7l + 5m – 2 = 0 & 1 = – 5l + 7m ⇒ – 5l + 7m – 1 = 0
∴ l/−5+14 = m/10+7 = 9/49+25 ⇒ l = 9/74 & m = 17/74
∴ from (i); we have
2 sinθ + cosθ = 9/74(7sinθ – 5cosθ) + 17/74(7 cosθ + 5 sinθ/)

(ii) Let I = ∫cosx+3sinx−5/3(1−sinx)−cosx dx
Numerator = l(deno) + m d/dx l(deno) + n
i.e. cos x + 3 sin x – 5 = l (3 – 3 sin x – cos x) l + m(- 3 cos x + sin x) + n …(1)
Coeff. of sin x : 3 = – 3l + m …(2)
Coeff. of cos x : 1 = – 1 – 3 m …(3)
Constant term; – 5 = 3 l + n …(4)
Multiply eqn. 2 by 3 + eqn. (3); we have
1 + 9 = – 9l – l ⇒ l = –10/10 = -1
∴ from (2); m = 3 + 3l = 3 – 3 = 0
∴ from (4); n = – 5 – 3l = – 5 + 3 = – 2
Thus from eqn. (1); we have
cos x + 3 sin x – 5 = – 1(3 – 3 sin x – cos x) + 0 (-3 cos x + sin x) – 2

Que-6: (i) ∫dx/4sin²x+5cos²x
(ii) ∫dx/2cos²x+sin²x
(iii) ∫dx/1+3sin²x
(iv) ∫dx/cos2x+3sin²x
(v) ∫dx/9cos²x+8sin²x+3
(vi) ∫dx/(2sinx+3cosx)²

Sol: (i) Let I = ∫dx/4sin²x+5cos²x
Divide Numerator and denomenator by cos² x; we have

(ii) Let I = ∫dx/2cos²x+sin²x
Divide numerator and deno by cos²x; we have
= ∫sec²x/2+tan²x dx
Put tan x = t ⇒ sec²x dx = dt
= ∫dt/2+t² = ∫dt/t²+(√2)²
= 1/√2 tan^-1(t/√2) + C
= 1/√2 tan^-1(tanx/√2) + C

(iii) Let I = ∫dx/1+3sin2x
Divide Numerator & deno by cos²x ; we have
= ∫sec²x/sec²x+3tan²x  dx= ∫sec²x/1+4tan²x dx
Put tan x = t ⇒ sec²x dx = dt

(iv) Let I = ∫dx/cos2x+3sin²x
= ∫dx/cos²x−sin²x+3sin²x
= ∫dx/cos²x+2sin²x
Divide Numerator and deno by cos²x; we have
= ∫sec²x/1+2tan²x dx
Put tan x = t ⇒ sec²x dx = dt

(v) Let I = ∫dx/9cos²x+8sin²x+3
Divide numerator and deno. by cos²x; we have

(vi) Let I = ∫dx/(2sinx+3cosx)²
Divide Numerator & denomirator by cos²x; we get

Que-7: (i) ∫dx/4+3sin2x
(ii) ∫dx/sin2x+4

Sol: (i) Let I = ∫dx/4+3sin2x
= ∫dx/4+6sinxcosx
Divide Numerator & denomirator by cos²x; we get

(ii) Let I = ∫dx/sin2x+4 = ∫dx/2sinx cosx+4
Divide numerator & denomirator by cos²x, we have

Que-8: ∫sin2x/( sin⁴x+cos⁴x)dx

Sol: Let I =∫sin2x/( sin⁴x+cos⁴x)dx;
Divide Numerator & denominator by cos⁴x; we have
= ∫2tanxsec²x/ tan⁴x+1 dx
put tan²x = t ⇒ 2 tan x sec²x dx = dt
= ∫dt/t²+1² = 1/1 tan^-1t + c
= tan^-1(tan²x) + C

–: End of Indefinite Integrals Class 12 OP Malhotra Exe-15G ISC Math Ch-15 Solution :–

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