Rd sharma solution class 1 chapter Indefinite integrals

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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