AP Inter 2nd Year Maths Exercise 7a Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7a

I. Find an anti derivative (or integral) of the following functions (1 to 5) by the method of inspection

Question 1.
sin 2x
Solution:
Method of Inspection:
We know that \(\frac{d}{d x}\)(cos2x) = -2sin2x dx
⇒ \(\frac{-1}{2} \frac{\mathrm{~d}}{\mathrm{dx}}\)(cos2x) = sin2x ⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)\) = sin 2x
By definition of integral, anti-derivative of sin2x is \(\frac{-1}{2}\)cos2x.

Question 2.
cos 3x
Solution:
Method of Inspection:
We know that \(\frac{d}{d x}\)(sin3x) = 3 cos3x dx
⇒ \(\frac{1}{3} \frac{\mathrm{~d}}{\mathrm{dx}}\)(sin3x)= cos3x ⇒ \(\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)\) = cos 3x
By definition of integral, anti-derivative of cos3x is \(\frac{1}{3}\)sin3x.

Question 3.
e^2x
Solution:
We know that \(\frac{d}{d x} e^{2 x}=e^{2 x} \frac{d}{d x}(2 x)=2 e^{2 x} \Rightarrow \frac{1}{2} \frac{d}{d x} e^{2 x}=e^{2 x} \Rightarrow \frac{d}{d x}\left(\frac{1}{2} e^{2 x}\right)\) = 2e^2x
∴ An antiderivative of e^2x is \(\frac{1}{2}\) e^2x.

Question 4.
(ax + b)²
Solution:
We know that \(\frac{d}{d x}\)(ax + b)³ = 3(ax + b)²\(\frac{d}{d x}\)(ax + b) = 3(ax + b)²a
⇒ \(\frac{1}{3 a} \frac{d}{d x}\)(ax + b)³ = (ax + b)² ⇒ \(\frac{d}{d x}\left[\frac{1}{3 a}(a x+b)^3\right]\) = (ax + b)²
∴ An antiderivative of (ax + b)² is \(\frac{1}{3a}\) (ax+b)³.

Question 5.
sin 2x – 4e^3x
Solution:
We know that \(\frac{d}{d x}\)(cos2x) = -2sin2x dx
⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)\) = sin 2x ……(i)
Again \(\frac{d}{d x}\)e^3x = 3e^3x
∴ \(\frac{d}{d x}\left(\frac{1}{3} e^{3 x}\right)=e^{3 x} \Rightarrow \frac{d}{d x}\left(\frac{-4}{3} e^{3 x}\right)\) = -4e^3x ………(ii)
Adding eqns. (i) and (ii) \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)+\frac{d}{d x}\left(\frac{-4}{3} e^{3 x}\right)\) = sin 2x – 4e^3x
⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\right)\) = sin 2x – 4e^3x
∴ An antiderivative of sin 2x – 4e^3x is \(\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\)

Question 6.
Find ∫(4e^3x + 1) dx
Solution:
∫(4e^3x + 1) dx = ∫4e^3x dx + ∫1 dx
= 4∫e^3x dx + x = 4\(\left(\frac{e^{3 x}}{3}\right)\) + x + c. [∵ ∫e^ax dx \(\frac{e^{a x}}{a}\) and ∫ 1 dx = x]

Question 7.
Find ∫ x²(1 – \(\frac{1}{x^2}\)) dx
Solution:
∫ x²(1 – \(\frac{1}{x^2}\)) dx = ∫(x² – \(\frac{x^2}{x^2}\)) dx = ∫ (x² – 1) dx
= ∫ x² dx – ∫ 1 dx = \(\frac{x^3}{3}\) – x + c. [∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) if n ≠ -1]

Question 8.
Find ∫ (ax² + bx + c) dx
Solution:
∫(ax² + bx + c) dx = ∫ ax² dx + ∫bx dx + ∫ x dx
= a ∫ x² dx + b ∫x¹ dx + c∫ 1 dx = a\(\frac{x^3}{3}\) + b\(\frac{x^2}{2}\) + cx + c₁
where c₁ is the constant of integration.

Question 9.
Find ∫ (2x² + e^x) dx
Solution:
∫(2x² + e^x)dx = ∫2x² dx + ∫ e^x dx
= 2∫x² dx + ∫e^x dx = 2\(\frac{x^{2+1}}{2+1}\) + e^x + c = \(\frac{2}{3}\)x³ + e^x + c.

Question 10.
Find \(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\) dx
Solution:
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\) dx
\(\int\left((\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2-2 \sqrt{x} \frac{1}{\sqrt{x}}\right)\) dx [∵ (a – b)² = a² – b² – 2ab]
= ∫(x + \(\frac{1}{x}\) – 2) dx = ∫x dx + ∫\(\frac{1}{x}\) dx – ∫2dx = \(\frac{x^2}{2}\) + log|x| – 2x + c. [∵∫2dx = 2∫1 dx = 2x]

Question 11.
Find \(\int \frac{x^3+5 x^2-4}{x^2} d x\)
Solution:
\(\int \frac{x^3+5 x^2-4}{x^2} d x=\int\left(\frac{x^3}{x^2}+\frac{5 x^2}{x^2}-\frac{4}{x^2}\right)\) dx
= ∫(x + 5 – 4x^-2) dx = ∫x¹ dx + ∫5 dx – ∫4x^-2 dx = \(\frac{x^2}{2}\) + 5 ∫1 dx – 4∫x^-2 dx
= \(\frac{x^2}{2}\) + 5x – 4\(\frac{x^{-2+1}}{-2+1}\) + c = \(\frac{x^2}{2}\) + 5x + \(\frac{4}{x}\) + c

Question 12.
Find \(\int \frac{x^3+3 x+4}{\sqrt{x}}\) dx
Solution:
\(\int \frac{x^3+3 x+4}{\sqrt{x}}\) dx = \(\int\left(\frac{x^3}{x^{1 / 2}}+\frac{3 x}{x^{1 / 2}}+\frac{4}{x^{1 / 2}}\right)\) dx
= ∫(x^3-1/2 + 3x^1-1/2 + 4x^-1/2) dx = ∫(x^5/2 + 3x^1/2 + 4x^-1/2) dx
= ∫x^5/2 dx + 3∫x^1/2 dx + 4∫x^{-1/2 dx}
= \(\frac{x^{5 / 2+1}}{\frac{5}{2}+1}+3 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+4 \frac{x^{-1 / 2+1}}{\frac{-1}{2}+1}+c=\frac{x^{7 / 2}}{\frac{7}{2}}+3 \frac{x^{3 / 2}}{\frac{3}{2}}+4 \frac{x^{1 / 2}}{\frac{1}{2}}+c\)
= \(\frac{2}{7}\) x^7/2 + 2x^3/2 + 8x^1/2 + c.

Question 13.
Find \(\int \frac{x^3-x^2+x-1}{x-1}\) dx
Solution:
\(\int \frac{x^3-x^2+x-1}{x-1} d x=\int \frac{x^2(x-1)+(x-1)}{x-1} d x=\int \frac{(x-1)\left(x^2+1\right)}{(x-1)} d x=\int\left(x^2+1\right) d x\)
= \(\int x^2 d x+\int 1 d x=\frac{x^{2+1}}{2+1}+x+c=\frac{x^3}{3}+x+c\)

Question 14.
Find ∫(1 – x)\(\sqrt{\mathbf{x}}\) dx
Solution:
∫(1 – x)\(\sqrt{\mathbf{x}}\) dx = \(\int(\sqrt{\mathrm{x}}-\mathrm{x} \sqrt{\mathrm{x}}) \mathrm{dx}=\int\left(\mathrm{x}^{1 / 2}-\mathrm{x}^1 \mathrm{x}^{1 / 2}\right) \mathrm{dx}=\int\left(\mathrm{x}^{1 / 2}-\mathrm{x}^{1+1 / 2}\right) \mathrm{dx}\)
= \(\int\left(x^{1 / 2}-x^{3 / 2}\right) d x=\frac{x^{1 / 2+1}}{\frac{1}{2}+1}-\frac{x^{3 / 2+1}}{\frac{3}{2}+1}+c=\frac{x^{3 / 2}}{\frac{3}{2}}-\frac{x^{5 / 2}}{\frac{5}{2}}+c=\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}+c \ldots\)

Question 15.
Find ∫\(\sqrt{x}\)(3x² + 2x + 3) dx
Solution:
∫\(\sqrt{x}\)(3x² + 2x + 3) dx = ∫x^1/2(3x² + 2x + 3) dx
= ∫(3x²x^1/2 + 2xx^1/2 + 3x^1/2) dx = ∫(3x^5/2 + 2x^3/2 + 3x^1/2) dx
= 3∫x^5/2 dx + 2∫x^3/2 dx + 3∫x^1/2 dx (∵\(2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}, 1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}\))
= \(3 \frac{x^{5 / 2+1}}{\frac{5}{2}+1}+2 \frac{x^{3 / 2+1}}{\frac{3}{2}+1}+3 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+c=3 \frac{x^{7 / 2}}{\frac{7}{2}}+2 \frac{x^{5 / 2}}{\frac{5}{2}}+3 \frac{x^{3 / 2}}{\frac{3}{2}}+c\)
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}+2 x^{3 / 2}+c\)

Question 16.
Find ∫(2x – 3cosx + e^x) dx
Solution:
∫(2x – 3cosx + e^x) dx = ∫2x dx – ∫3cos x dx + ∫e^x dx
= 2 ∫x¹ dx – 3∫cosx dx + e^x dx = 2\(\frac{x^2}{2}\) – 3sin x + e^x + c = x² – 3sin x + e^x + c

Question 17.
Find ∫(2x² – 3sin x + 5\(\sqrt{x}\)) dx
Solution:
∫(2x² – 3sin x + 5\(\sqrt{x}\)) dx = 2∫x² dx – 3∫sinx dx + 5∫x^1/2 dx
= \(2 \frac{x^{2+1}}{2+1}-3(-\cos x)+5 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+c=2 \frac{x^3}{3}+3 \cos x+5 \frac{x^{3 / 2}}{\frac{3}{2}}+c\)
= \(\frac{2}{3}\)x³ + 3cos x + \(\frac{10}{3}\)x^3/2 + c.

Question 18.
Find ∫secx(secx + tanx) dx
Solution:
∫secx(secx + tanx) dx = ∫(sec² x + sec x tan x) dx
= ∫sec² x dx + ∫secx tanx dx = tan x + sec x + c.

Question 19.
Find \(\int \frac{\sec ^2 x}{\operatorname{cosec}^2 x}\) dx
Solution:
\(\int \frac{\sec ^2 x}{\operatorname{cosec}^2 x} d x=\int \frac{\left(\frac{1}{\cos ^2 x}\right)}{\left(\frac{1}{\sin ^2 x}\right)} d x=\int \frac{\sin ^2 x}{\cos ^2 x} d x\)
= ∫tan²x dx = ∫(sec² x – 1)dx = tan x – x + c (∵ sec²x – tan² x = 1 ⇒ sec²x – 1 = tan²x)

Question 20.
Find \(\int \frac{2-3 \sin x}{\cos ^2 x} d x\)
Solution:
\(\int \frac{2-3 \sin x}{\cos ^2 x} d x=\int\left(\frac{2}{\cos ^2 x}-\frac{3 \sin x}{\cos ^2 x}\right) d x\)
= \(\int\left(2 \sec ^2 x-\frac{3 \sin x}{\cos x \cos x}\right) d x\) = ∫(2 sec² x – 3tan x sec x) dx
= 2∫sec² xdx – 3∫secx tanxdx = 2 tanx – 3 secx + c

Question 21.
Find the integral of \(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
Solution:
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}\)
= \(\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}=\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b}\)
⇒ \(\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x\)
= \(\frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+C\)

Question 22.
Find the integral of \(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
Solution:
Given integral is \(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}=\frac{e^{4 \log x}\left(e^{\log x}-1\right)}{e^{2 \log x}\left(e^{\log x}-1\right)}\) = e^{2log x} = e^{log x2} = x²
∴ \(\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int x^2 d x=\frac{x^3}{3}+C\)