Indefinite Integrals Class 12 OP Malhotra Exe-15A ISC Maths Solutions Ch-15 Solutions. In this article you would learn about integration of the standard forms ∫dx/x²±a² and ∫dx/a²-x² . 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-15A ISC Maths Solutions Ch-15
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-15 | Indefinite Integrals |
| Writer | OP Malhotra |
| Exe-15(a) | integration of the standard forms ∫dx/x²±a² and ∫dx/a²-x² |
Integration of the standard forms ∫dx/x²±a² and ∫dx/a²-x²
Indefinite Integrals Class 12 OP Malhotra Exe-15A Solutions
Que-1: (i) ∫1/x²+36 dx
(ii) ∫1+x²/4 dx
(iii) ∫dx/50+2x²
Sol: (i) ∫1/x²+36 dx = ∫1/x²+6² dx
= 1/6 tan-1 x/6 + C
[∵ ∫dx/x²+a² = 1/a tan-1 x/a + C]
(ii) ∫1+x²/4 dx = ∫4/2²+x² dx
= 4/2 tan-1 x/2 = 2 tan-1 x/2 + C
(iii) ∫dx/50+2x² = 1/2∫dx/x¹+5²
= 1/2 × 1/5 tan-1 (x/5) + C
= 1/10 tan-1 (x/5) + C
Que-2: (i)∫dx/x²-4
(ii) ∫dx/9x²-16
(iii) ∫dy/25-16y²
(iv) ∫dx/18-2x²
Sol:
Que-3: (i) ∫dx/(x+2)²+1
(ii) ∫dx/1+2(x+2)²
Sol: (i) Let I =∫dx/(x+2)²+1;
Put x + 2 = t ⇒ dx = dt
= ∫dt/t²+1² = tan-1(t/1) + C
= tan-1(x + 1) + C
(ii) Let I =∫dx/1+2(x+2)²;
Put x + 2 = t ⇒ dx = dt
Que-4: (i)∫3x²/x6+1 dx
(ii) ∫x/1+x4 dx
(iii) ∫cosx/1+sin²x dx
(iv) ∫ex/1+e2x dx
(v) ∫dx/ex+e-x
(vi) ∫e-x/16+9e-2x dx
(vii) ∫√ex-1 dx
Sol: (i) Let I =∫3x²/x6+1 dx = ∫3x²(x³)²+1
Put x³ = t ⇒ 3x²dx = dt
∴ I = ∫dt/t²+1² = 1/2 tan-1 t + C
= tan-1 x/3 + C
(ii) Let I =∫x/1+x4 dx;
Put x² = t ⇒ 2x dx = dt
= ∫dt/2(1+t²) = 1/2 tan-1 t + C
= 1/2 tan-1 (x²) + C
(iii) Let I = ∫cosx/1+sin²x dx;
Put sin x = t ⇒ cos x dx = dt
= ∫dt/1+t² = tan-1t + C
= tan-1(sin x) + C
(iv) Let I =∫ex/1+e2x dx;
Put ex = t ⇒ ex dx = dt
∴ I = ∫dt/1+t² = 1/1 tan-1t + C
= tan-1(ex) + C
(v) Let I = ∫dx/ex+e-x = ∫ex/e2x+1 dx;
Put ex = t ⇒ ex dx = dt
= ∫dt/t²+1 = tan-1t + C
= tan-1(ex) + C
(vi) Let I =∫e-x/16+9e-2x dx ;
Put ex = t ⇒ ex dx = dt
= ∫−dt/16+9t² = –1/9∫dt/t²+(4/3)²
= –1/9×1/(4/3) tan-1(t/(4/3)) + C
= –1/12 tan-1 (3e-x/4) + C
(vii) Let I =∫√ex-1 dx;
Put√ex-1= t ⇒ ex-1 = t²
⇒ex = t² + 1 ⇒ex dx = 2t dt
⇒ dx = 2t/t²+1 dt
∴ I = ∫2t²/t²+1 dt
= 2∫[1− 1/t²+1]dt
= 2[t−tan-1t]+C
= [√ex-1 − tan-1 √ex-1] + C
Que-5: (i) ∫x²−1/x²+4 dx
(ii) ∫x4/x²+1 dx
(iii) ∫3x5/1+x1/2 dx
(iv) ∫dx/2+cosx
Sol:
(ii) Let I =∫x4/x²+1 dx
= ∫x4-1+1/x²+1 dx
= ∫x4-1/x²+1 dx + ∫dx/x²+1
= ∫(x²−1)dx + ∫dx/x²+1²
= x³/3 – x + tan-1x + C
(iii) Let I = ∫3x5/1+x1/2 dx;
Put x6 = t ⇒ 6x5 dx = dt
= 3/6∫dx/1+t² = 1/2 tan-1t + C
= 1/2 tan-1 x6 + C
(iv) Let I =∫dx/2+cosx
Put tan x² = t ⇒ sec² x/2 1/2 dx = dt
⇒ dx = 2/1+t² dt
Que-6: (i) ∫x/x4−a4 dx
(ii) ∫x²/a6−x6 dx
(iii) ∫x/1−x4 dx
Sol: (i) Let I =∫x/x4−a4 dx;
Put x² = t ⇒ 2x dx = dt
(ii) Let I =∫x²/a6−x6 dx;
Put x3 = t ⇒ 3x2dx = dt
(iii) Let I =∫x/1−x4 dx;
Put x² = t ⇒ 2x dx = dt
= 1/2∫dt/1−t² = 1/2∫dt/1²−t²
Que-7: (i) ∫x³+x/x4−9 dx
(ii) ∫cosx/4−sin²x dx
(iii) ∫log(2+x²) dx
(iv) ∫x²−4/x4+16 dx
Sol: (i) Let I =∫x³+x/x4−9 dx = ∫x(x²+1)/x4−9 dx
= ∫x³/x4−9 dx + ∫x/x4−9 dx
= 1/4 ∫4x³/x4−9 dx +∫dt/2(t²−9) [Put x² = t ⇒ 2x dx = dt]
(ii) Let I =∫cosx/4−sin²x dx;
Put sin x = t ⇒ cos x dx = dt
= ∫dt/4−t² = ∫dt/2²−t²
(iii) Let I =∫log(2+x²) dx;
= log(2 + x²) x – ∫2x/2+x² xdx
= x log(2 + x²) – 2∫2+x²−2/2+x² dx
= x log(2+x²) – 2∫[1−2/2+x²]dx
= x log(2 + x²) – 2x + 4∫dx/x²+(√2)²
= x log(2 + x²) – 2x + 4∫dx/x²+(√2)²
= x log(2 + x²) – 2x + 4/√2 tan-1(x/√2) + C
(iv) Let I =∫x²−4/x4+16 dx ;
Divide numerator & deno. by x²; we have
Que-8: (i) ∫x²/x4+1 dx
(ii)∫dx/x4+1
(iii) ∫dx/x4+16
(iv) ∫x²/x4+16 dx
(v) ∫(x²+1)/x4+7x²+1 dx
(vi) ∫x²+4/x4+16 dx
Sol: (i) Let I =∫x²/x4+1 dx = 1/2∫2x²/x4+1 dx
= 1/2∫(x²+1)+(x²−1)x4+1
= 1/2∫x²+1/x4+1 dx + 1/2∫x²−1/x4+1 dx
= 1/2 I1 + 1/2 I2 …(1)
where I1 = ∫x²+1/x4+1 dx;
divide Numerator & deno by x²; we have
(ii) Let I =∫dx/x4+1 = 1/2∫2/x4+1
= 1/2 ∫(x²+1)−(x²−1)x4+1 dx
= 1/2 ∫x²+1/x4+1 dx – 1/2 ∫x²−1/x4+1 dx
= 1/2 I1 + 1/2 I2 …(1)
Where I1 = ∫x²+1/x4+1 dx;
divide numerator & deno by x² ; we have
Que-9: (i) ∫x²+1/x4+x²+1 dx
(ii) ∫x²+9/x4−2x²+81 dx
(iii) ∫x²−8/x4+7x²+64 dx
Sol: (i) Let I = ∫x²+1/x4+x²+1 dx
Divide numerator and deno by x2; we have
Que-10: i) ∫x²/x4+x²+1 dx
(ii) ∫dx/x4+x²+1
Sol: (i) Let I =∫x²/x4+x²+1 dx
= 1/2∫2x²/x4+x²+1 dx
= 1/2∫(x²+1)+(x²−1)/x4+x²+1 dx
= 1/2∫x²+1/x4+x²+1 dx + 1/2 ∫x²−1/x4+x²+1 dx
= 1/2I1 – 1/2I2 …(1)
Where I1 = ∫x²+1/x4+x²+1 dx;
Divide numerator and deno. by x2
Put x – 1/x = t ⇒ (1+1/x²)dx = dt
on squaring both sides; we have
(x−1/x)² = t² ⇒ x² + 1/x² – 2 = t²
⇒ x² + 1/x² = t² + 2
Divide numerator and deno by x2; we have
(ii) Let I = ∫dx/x4+x²+1
= 1/2∫2dx/x4+x²+1
= 1/2∫(x²+1)−(x²−1)x4+x²+1 dx
= 1/2I1 – 1/2I2
= 1/2 ∫x²+1/x4+x²+1d x + 12∫x²−1/x4+x²+1 dx
where I1 = ∫x²+1/x4+x²+1 dx;
Divide numerator and demo. by x²; we have
Que-11: ∫(x−1)²/x4+x²+1 dx
Sol: Let I = ∫(x−1)²/x4+x²+1 dx = ∫x²−2x+1/x4+x²+1 dx
= ∫x²+1/x4+x²+1 dx = ∫2x/x4+x²+1 dx
= I1–I2…(1)
∴ I2= ∫x²+1/x4+x²+1 dx;
divide Num & Demo by x²; we have
Put x – 1/x = t ⇒ (1+1/x²)dx = dt
on squaring both sides; we have
(x−1/x)² = t²
⇒ x² + 1/x² – 2 = t²
⇒ x² + 1/x² = t² + 2
Putting eqn. (2) & eqn. (3) in eqn.(1); we have
I = 1/√3 tan-1 (x²−1/x√3) – 2/√3 tan-1 (2x²+1/√3) + C
where C1 – C2 = C
–: End of Indefinite Integrals Class 12 OP Malhotra Exe-15A ISC Math Ch-15 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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