Balbharti 12th Maharashtra State Board Maths Solutions Book

Pdf Chapter 3 Indefinite Integration Ex 3.3 Questions and Answers.

I. Evaluate the following:

Question 1.
∫x² log x dx
Solution:


Question 2.
∫x² sin 3x dx
Solution:





Question 3.
∫x tan^-1 x dx
Solution:


Question 4.
∫x² tan^-1 x dx
Solution:


Question 5.
∫x³ tan^-1 x dx
Solution:
Let I = ∫x³ tan^-1 x dx
= ∫(tan^-1 x) . x³ dx


Question 6.
∫(log x)² dx
Solution:
Let I = ∫(log x)² dx
Put log x = t



Question 7.
∫sec³ x dx
Solution:
Let I = ∫sec³ x dx
= ∫sec x sec² x dx
= sec x ∫sec² x dx – ∫[\(\frac{d}{d x}\)(sec x) ∫sec² x dx] dx
= sec x tan x – ∫(sec x tan x)(tan x) dx
= sec x tan x – ∫sec x tan² x dx
= sec x tan x – ∫sec x (sec² x – 1) dx
= sec x tan x – ∫sec³ x dx + ∫sec x dx
∴ I = sec x tan x – I + log|sec x + tan x|
∴ 2I = sec x tan x + log|sec x + tan x|
∴ I = \(\frac{1}{2}\) [sec x tan x + log|sec x + tan x|] + c.



Question 8.
∫x . sin² x dx
Solution:


Question 9.
∫x³ log x dx
Solution:


Question 10.
∫e^2x cos 3x dx
Solution:



Question 11.
∫x sin^-1 x dx
Solution:





Question 12.
∫x² cos^-1 x dx
Solution:


Question 13.
\(\int \frac{\log (\log x)}{x} d x\)
Solution:

= t(log t – 1) + c
= (log x) . [log(log x) – 1] + c.

Question 14.
\(\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Solution:


Question 15.
∫cos√x dx
Solution:
Let I = ∫cos√x dx
Put √x = t
∴ x = t²
∴ dx = 2t dt
∴ I = ∫(cos t) 2t dt
= ∫2t cos t dt
= 2t ∫cos t dt – ∫[\(\frac{d}{d t}\)(2t) ∫cos t dt]dt
= 2t sin t – ∫2 sin t dt
= 2t sin t + 2 cos t + c
= 2[√x sin√x + cos√x] + c.



Question 16.
∫sin θ . log(cos θ) dθ
Solution:
Let I = ∫sin θ . log (cos θ) dθ
= ∫log(cos θ) . sin θ dθ
Put cos θ = t
∴ -sin θ dθ = dt
∴ sin θ dθ = -dt

= -t log t + t + c
= -cos θ . log(cos θ) + cos θ + c
= -cos θ [log(cos θ) – 1] + c.

Question 17.
∫x cos³ x dx
Solution:
cos 3x = 4 cos³ x – 3 cos x
∴ cos³ x + 3 cos x = 4cos3x
∴ cos³ x = \(\frac{1}{4}\) cos 3x + \(\frac{3}{4}\) cos x


Question 18.
\(\int \frac{\sin (\log x)^{2}}{x} \cdot \log x d x\)
Solution:




Question 19.
\(\int \frac{\log x}{x} d x\)
Solution:
Let I = \(\int \frac{\log x}{x} d x\)
Put log x = t
\(\frac{1}{x}\) dx = dt
∴ I = ∫t dt
= \(\frac{1}{2}\) t² + c
= \(\frac{1}{2}\) (log x)² + c

Question 20.
∫x sin 2x cos 5x dx.
Solution:
Let I = ∫x sin 2x cos 5x dx
sin 2x cos 5x = \(\frac{1}{2}\)[2 sin 2x cos 5x]
= \(\frac{1}{2}\) [sin(2x + 5x) + sin(2x – 5x)]
= \(\frac{1}{2}\) [sin 7x – sin 3x]
∴ ∫sin 2x cos 5x dx = \(\frac{1}{2}\) [∫sin 7x dx – ∫sin 3x dx]


Question 21.
\(\int \cos (\sqrt[3]{x}) d x\)
Solution:
Let I = \(\int \cos (\sqrt[3]{x}) d x\)


II. Integrate the following functions w.r.t. x:

Question 1.
e^2x sin 3x
Solution:





Question 2.
e^-x cos 2x
Solution:



Question 3.
sin(log x)
Solution:


Question 4.
\(\sqrt{5 x^{2}+3}\)
Solution:
Let I = \(\sqrt{5 x^{2}+3}\) dx


Question 5.
\(x^{2} \sqrt{a^{2}-x^{6}}\)
Solution:


Question 6.
\(\sqrt{(x-3)(7-x)}\)
Solution:




Question 7.
\(\sqrt{4^{x}\left(4^{x}+4\right)}\)
Solution:


Question 8.
(x + 1) \(\sqrt{2 x^{2}+3}\)
Solution:
Let I = ∫(x + 1) \(\sqrt{2 x^{2}+3}\) dx
Let x + 1 = A[\(\frac{d}{d x}\)(2x² + 3)] + B
= A(4x) + B
= 4Ax + B
Comparing the coefficients of x and constant term on both the sides, we get
4A = 1, B = 1
∴ A = \(\frac{1}{4}\), B = 1



Question 9.
\(x \sqrt{5-4 x-x^{2}}\)
Solution:
Let I = ∫\(x \sqrt{5-4 x-x^{2}}\) dx
Let x = A[\(\frac{d}{d x}\)(5 – 4x – x²)] + B
= A[-4 – 2x] + B
= -2Ax + (B – 4A)
Comparing the coefficients of x and the constant term on both sides, we get
-2A = 1, B – 4A = 0





Question 10.
\(\sec ^{2} x \sqrt{\tan ^{2} x+\tan x-7}\)
Solution:



Question 11.
\(\sqrt{x^{2}+2 x+5}\)
Solution:


Question 12.
\(\sqrt{2 x^{2}+3 x+4}\)
Solution:



III. Integrate the following functions w.r.t. x:

Question 1.
[2 + cot x – cosec² x] e^x
Solution:
Let I = ∫e^x [2 + cot x – cosec² x] dx
Put f(x) = 2 + cot x
∴ f'(x) = \(\frac{d}{d x}\)(2 + cot x)
= \(\frac{d}{d x}\)(2) + \(\frac{d}{d x}\)(cot x)
= 0 – cosec² x
= -cosec² x
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x (2 + cot x) + c.



Question 2.
\(\left(\frac{1+\sin x}{1+\cos x}\right) e^{x}\)
Solution:


Question 3.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution:
Let I = ∫\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Let f(x) = \(\frac{1}{x}\), f'(x) = \(-\frac{1}{x^{2}}\)
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x . \(\frac{1}{x}\) + c

Question 4.
\(\left[\frac{x}{(x+1)^{2}}\right] e^{x}\)
Solution:


Question 5.
\(\frac{e^{x}}{x}\) . [x(log x)² + 2 log x]
Solution:




Question 6.
\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Solution:
Let I = ∫\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)


Question 7.
\(e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right]\)
Solution:


Question 8.
log(1 + x)^(1+x)
Solution :
Let I = ∫log(1 + x)^(1+x) dx




Question 9.
cosec (log x)[1 – cot(log x)]
Solution:
Let I = ∫cosec (log x)[1 – cot(log x)] dx
Put log x = t
x = e^t
dx = e^t dt
I = ∫cosec t (1 – cot t). e^t dt
= ∫e^t [cosec t – cosec t cot t] dt
= ∫e^t [cosec t + \(\frac{d}{d t}\) (cosec t)] dt
= e^t cosec t + c ….. [∵ e^t [f(t) +f'(t) ] dt = e^t f(t) + c ]
= x . cosec(log x) + c.