Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(f) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(f)

I. Evaluate the following integrals.

Question 1.

∫e^{x}(1 + x²) dx

Solution:

∫e^{x}(1 + x²) dx = ∫e^{x} + ∫x² . e^{x} dx

= e^{x} + (x².e^{x} – 2 ∫x.e^{x}dx)

= e^{x} + x².e^{x} – 2(x.e^{x} – ∫e^{x} dx

= e^{x} + x².e^{x} – 2x. e^{x} + 2e^{x} + C

= e^{x}(x² – 2x + 3) + C

Question 2.

∫x² e^{-3x} dx

Solution:

Question 3.

∫x³ e^{ax} dx

Solution:

II.

Question 1.

Show that

∫x^{n}.e^{-x} dx = -x^{n} e^{-x} + n∫x^{n-1}.e^{-x} dx

Solution:

∫x^{n}.e^{-x} dx = \(\frac{x^ne^{-x}}{(-1)}\) + ∫e^{-x}. nx^{n-1} dx

= -x^{n} . e^{-x} + n∫x^{n-1}e^{-x} dx

Question 2.

If I_{n} = ∫cos^{n} x dx, than show that

I_{n} = \(\frac{1}{n}\) cos^{n-1} x sin x + \(\frac{n-1}{n}\) I_{n-2}.

Solution:

I_{n} = ∫cos^{n} x dx = ∫cos^{n-1} x. cos x dx

= cos^{n-1}x . sin x – ∫sin x.(n – 1)cos^{n-2}x (-sin x)dx

= cos^{n-1}x . sin x + (n – 1) ∫cos^{n-2}x(1 – cos² x)dx

= cos^{n-1}x . sin x + (n – 1)I_{n-2} – (n – 1) I_{n}

∴ I_{n}(1 + n – 1) = cos^{n-1}x. sin x + (n – 1) I_{n-2}

I_{n} = \(\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n}\)I_{n-2}

III.

Question 1.

Obtain reduction formula for I_{n} = ∫cot^{n} x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot^{4} x dx.

Solution:

I_{n} = ∫cot^{n} dx = ∫cot^{n-2} x . cot² x dx

= ∫cot^{n-2} x . (cosec² x – 1) dx

= ∫cot^{n-2} x . cosec² x dx – I_{n-2}

= – \(\frac{cot^{n-1}x}{n-1}\) – I_{n-2}

n = 4 ⇒ I_{4} = –\(\frac{cot^{3}x}{3}\) – I_{2}

n = 2 ⇒ I_{2} = – cot x – I_{0} where I_{0} = ∫dx = x

I_{2} = – cot x – x

I_{4} = –\(\frac{cot^{3}x}{3}\) – (-cot x – x) + C

= –\(\frac{cot^{3}x}{3}\) + cot x + x + C

Question 2.

Obtain the reduction formula for I_{n} = ∫cosec^{n} x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec^{5} x dx.

Solution:

I_{n} = ∫cosec^{n} x dx

= cosec^{n-2} x(-cot x) + ∫cot x. (n – 2)cosec^{n-3} x. (cot x)dx

= -cosec^{n-2}x. cot x + (n – 2)∫cosec^{n-2}x. (cosec² x – 1)dx

Question 3.

If I_{m, n} = ∫sin^{m} xcos^{n} xdx, then show that

for a positive integer m ≥ 2.

Solution:

I_{m, n} = ∫(sin^{m} x) (cos^{n} x) dx

= ∫sin^{m-1} (x).(cos x)^{n}. sin x dx

= ∫sin^{m-1} (x)(cos x)^{n}(-sin x) dx

Question 4.

Evaluate ∫sin^{5} x cos^{4} x dx

Solution:

Reduction formula

I_{m, n} = \(\frac{-\sin ^{m-1} x \cdot \cos ^{n+1} x}{m+n}+\frac{m-1}{m+n}\).I_{m-2, n
}

Question 5.

If I_{n} = ∫(log x)^{n} dx, then show that I_{n} = x(log x)^{n} – nI_{n-1}, and hence find

∫(log x)^{4} dx.

Solution:

I_{n} = ∫(log x)^{n} dx

= (log x)^{n}. x – ∫x . n . (log x)^{n-1} . \(\frac{1}{x}\)dx

= x.(log x)^{n} – n∫(log x)^{n-1} dx

= x(log x)^{n} – n . I_{n-1}

I_{4} = x(log x)^{4} – 4 . I_{3}

I_{3} = x(log x)³ – 3 . I_{2}

I_{2} = x(log x)² – 2 . I_{1}

I_{1} = x log x I_{0} where I_{0} = ∫dx = x

I_{1} = x log x – x

I_{2} = (x (log x)² – 2x log x = 2x

I_{3} = x(log x)³ – 3(x (log x)² – 2x log x + 2x)

= x . (log x)³ – 3x (log x)² + 6x(log x) – 6x

I_{4} = x(log x)^{4} – 4[x . (log x)³ – 3x(log x)² + 6x(log x) -6x] + C

= x[(log x)^{4} – 4(log x)³ + 12(log x)² – 24(log x) + 24] + C