Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 7 Integrals Ex 7.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.6

Integrate the following functions:

Question 1.

x sin x

Solution:

Let I = âˆ«x sin x dx

We know that âˆ«uv dx = uâˆ«v dx – âˆ«(u’ âˆ«v dx)dx

Put u = x and v = sin x.

âˆ´ I = x âˆ«sin x dx = âˆ«[(\(\frac{d}{dx}\) x) âˆ«sin x dx]dx

= – x cos x – âˆ«1. (- cos x)dx

= – x cos x + âˆ«cos x dx = – x cos x + sin x + C.

Question 2.

x sin 3x

Solution:

Let I = âˆ«xsin 3x dx. Taking u = x and v = sin 3x.

Question 3.

x^{2}e^{x}

Solution:

Let I = âˆ«x^{2}e^{x} dx, Put u = x^{2} and v = e^{x}dx.

Taking x as I function, integrating again, we get

Question 4.

x log x

Solution:

Let I = âˆ«xlog x dx, Put u = log x and v = x.

Question 5.

x log 2x

Solution:

Let I = âˆ«xlog 2x dx = âˆ«(log 2x).x dx

Question 6.

x^{2}log x

Solution:

Question 7.

x sin^{-1}x

Solution:

Put x = sinÎ¸ so that dx = cosÎ¸dÎ¸.

Question 8.

x tan^{-1}x

Solution:

Let I = âˆ«x tan^{-1} x dx. Put x = tan^{-1}x and v = x.

Question 9.

x cos^{-1}x

Solution:

Let I = âˆ«xcos^{-1} x dx = âˆ«cos^{-1} x.x dx

Putting x = cosÎ¸ so that dx = – sinÎ¸ dÎ¸.

Question 10.

(sin^{-1}x)^{2}

Solution:

Let I = âˆ«(sin^{-1} x^{2})dx.

Put sin^{-1}x = Î¸ â‡’ x = sin Î¸ â‡’ dx = cosÎ¸ dÎ¸.

Question 11.

\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)

Solution:

Question 12.

x sec^{2}x

Solution:

Question 13.

tan^{-1}x

Solution:

Question 14.

x(log x)^{2}

Solution:

Question 15.

(x^{2} + 1) log x

Solution:

Question 16.

e^{x}(sin x + cos x)

Solution:

Let I = âˆ«e^{x}(sin x + cos x) dx.

Put e^{x}sin x = t so that (e^{x}.cos x + sin x.e^{x})dx = dt.

â‡’ e^{x}(sin x + cos x) dx = dt

âˆ´ I = âˆ«dt = t + C = e^{x}sinx + C.

Question 17.

\(\frac{x e^{x}}{(1+x)^{2}}\)

Solution:

Question 18.

e^{x}(\(\frac{1+sinx}{1+cosx}\))

Solution:

Taking tan \(\frac{x}{2}\) as I function and e^{x} as II function,

integrating âˆ«e^{x} tan \(\frac{x}{2}\) dx by parts, we get

Question 19.

e^{x}(\(\frac{1}{x}\) – \(\frac{1}{x^{2}}\))

Solution:

Question 20.

\(\frac{(x-3) e^{x}}{(x-1)^{3}}\)

Solution:

Question 21.

e^{2x}sinx

Solution:

Question 22.

sin^{-1}(\(\frac{2 x}{1+x^{2}}\))

Solution:

Choose the correct answers in the following questions from 23 and 24:

Question 23.

âˆ« x^{2} e^{x3} dx equals

(A) \(\frac{1}{3}\) e^{x3} + C

(B) \(\frac{1}{3}\) e^{x2} + C

(C) \(\frac{1}{2}\) e^{x3} + C

(D) \(\frac{1}{2}\) e^{x2} + C

Solution:

âˆ´ Part (A) is the correct answer.

Question 24.

âˆ« e^{x} secx(1 + tan x) dx equals

(A) e^{x} cosx + C

(B) e^{x} secx + C

(C) e^{x} sinx + C

(D) e^{x} tanx + C

Solution:

Taking sec x as I function and e^{x} as II function, integrating by parts

I_{1} = (sec x) âˆ«e^{x} dx – âˆ«(sec x tan x âˆ«e^{x} dx)dx

= (sec x)e^{x} – âˆ«e^{x} sec x tan x dx

Putting this value in (1), we get

I = I_{1} + âˆ«e^{x} sec x tan x dx

= (sec x)e^{x} – âˆ«e^{x} sec x tan x dx

Putting this value in (1), we get

I = I_{1} + âˆ«e^{x} sec x tan x dx

= (sec x)e^{x} – âˆ«e^{x} sec x tan x dx + âˆ«e^{x} sec x tan x dx + C

= e^{x} sec x + C.

âˆ´ Part (B) is the correct answer.