Rd sharma solution class 1 chapter Indefinite integrals

Leave a Comment / class 12, home / By Tanishk

Exercise 19.1

Question 1

Evaluate each of the following integrals:

(i) $\int{x^4~dx}$

Sol :

$=\int{x^4~dx}$

$=\dfrac{x^{4+1}}{4+1}+C$

$=\dfrac{x^{5}}{5}+C$

(ii) $\int{x^{\dfrac{5}{4}}~dx}$

Sol :

$=\dfrac{x^{\frac{5}{4}+1}}{\frac{5}{4}+1}$

$=\dfrac{4}{9}x^{\frac{9}{4}}+C$

(iii) $\int\dfrac{1}{x^5}dx$

Sol :

$=\dfrac{x^{-5+1}}{-5+1}+C$

$=-\dfrac{1}{4}x^{-4}+C$

$=-\dfrac{1}{4x^4}+C$

(iv) $\int\dfrac{1}{x^{3/2}}dx$

Sol :

$=\int{x^{-3/2}}dx$

$=\bigg[\dfrac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\bigg]+C$

$=\bigg[\dfrac{x^{-\frac{1}{2}}}{-\frac{1}{2}}\bigg]+C$

$=-\dfrac{2}{\sqrt{x}}+C$

(v) $\int3^xdx$

Sol :

$=\dfrac{3^x}{log_ea}+C$

(vi) $\int\dfrac{1}{\sqrt[3]{x^2}}dx$

Sol :

$=\int\dfrac{dx}{x^{2/3}}$

$=\int x^{-2/3}dx$

$=\dfrac{x^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}$

$=3x^{\dfrac{1}{3}}+C$

(vii) $\int3^{2log_3x}dx$

Sol :

$=\int 3^{log_3x^2}dx$

$=\int x^2dx$

$=\dfrac{x^3}{3}+C$

(viii) $\int{log_xx}dx$

Sol :

$=\int1.dx$

$=x+C$

Question 2

Evaluate :

(i) $\int{\dfrac{1+cos~2x}{2}}dx$

Sol :

$=\int{\dfrac{1+cos~2x}{2}}dx$

$=\int{\dfrac{2cos^2x}{2}}dx$

$[Since,~1+cos~2x=2cos^2x]$

$=\int cos~x~dx$

$=sin~x+C$

(ii) $\int{\dfrac{1-cos~2x}{2}}dx$

Sol :

$=\int{\dfrac{1-cos~2x}{2}}dx$

$=\int{\dfrac{2sin^2x}{2}}dx$

$[Since,~1-cos~2x=2sin^2x]$

$=\int sin~x~dx$

$=-cos~x+C$

Question 3

Evaluate : $\int \dfrac{e^{6~log_ex}-e^{5~log_ex}}{e^{4~log_ex}-e^{3~log_ex}}dx$

Sol :

$=\int \dfrac{e^{6~log_ex}-e^{5~log_ex}}{e^{4~log_ex}-e^{3~log_ex}}dx$

$=\int \dfrac{e^{log_ex^6}-e^{log_ex^5}}{e^{log_ex^4}-e^{log_ex^3}}dx$

$=\int\dfrac{x^6-x^5}{x^4-x^3}dx$

$=\int\dfrac{x^5(x-1)}{x^3(x-1)}dx$

$=\int x^2dx$

$=\dfrac{x^3}{3}+C$

Question 4

Evaluate : $\int \dfrac{1}{a^xb^x}dx$

Sol :

$=\int \dfrac{1}{a^xb^x}dx$

$=\int (a^{-x}~b^{-x})dx$

$=\int (ab)^{-x}dx$

$=\dfrac{(ab)^{-x}}{-log_e{ab}}dx+C$

$=\dfrac{a^{-x}b^{-x}}{-log_e{ab}}dx+C$

Question 5

Evaluate :

(i) $\int \dfrac{cos~2x+2sin^2~x}{sin^2~x}dx$

Sol :

$=\int \dfrac{cos~2x+2sin^2~x}{sin^2~x}dx$

$=\int \dfrac{1-2sin^2~x+2sin^2~x}{sin^2~x}dx$

$[Since,~cos~2x=1-2sin^2~x]$

$=\int \dfrac{1}{sin^2~x}dx$

$=\int cosec^2~x~dx$

$=-cot~x+C$

(ii) $\int \dfrac{2cos^2~x-cos~2x}{cos^2~x}dx$

Sol :

$=\int \dfrac{2cos^2~x-(2cos^2~x-1)}{cos^2~x}dx$

$[Since,~cos~2x=2cos^2~x-1]$

$=\int \dfrac{1}{cos^2~x}dx$

$=\int sec^2~x~dx$

$=tan~x+C$

Question 6

Evaluate :

$\int \dfrac{e^{log_e~\sqrt{x}}}{x}dx$

Sol :

$=\int \dfrac{e^{log_e~\sqrt{x}}}{x}dx$

$=\int \dfrac{\sqrt{x}}{x}dx$

$=\int \dfrac{1}{\sqrt{x}}dx$

$=\int x^{-\frac{1}{2}}~dx$

$=\Bigg[\dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+C$

$=2\sqrt{x}+C$

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