Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.9

Solutions for NCERT Class 12 Maths Chapter 7 Exercise 7.9 (Integrals) are given in the article today and the main formula used while solving this exercise is shown below:

If\, \int f(x)=F(x)\,then\\ \int_{a}^{b} f(x) \,dx=[ \,F(x)] \,_{a}^{b}=F(b)-F(a)

Chapter 7– Integrals Exercise 7.9

Question 1: \int_{-1}^{1} (x+1) \,dx

Answer:

Let \,I = \int_{-1}^{1} (x+1) \,dx

\int(x+1)dx = \frac{x^2}{2}+x=F(x)

Using second fundamental theorem of calculus

I =F(1)-F(-1)

=(\frac{1}{2}+1)-(\frac{1}{2}-1)

=\frac{1}{2}+1-\frac{1}{2}+1

= 2

Question 2: \int_{2}^{3} \frac{1}{x} \,dx

Answer:

Let I=\int_{2}^{3} \frac{1}{x} \,dx

\int \frac{1}{x} dx = log|x|=F(x)

Using second fundamental theorem of calculus

I = F(3) - F(2)

= log|3| - log|2|

= log(3/2)

Question 3: \int_{1}^{2} (4x^3-5x^2+6x+9) \,dx

Answer:

Let\,I = \int_{1}^{2} (4x^3-5x^2+6x+9) \,dx

\int (4x^3-5x^2+6x+9) \,dx =4(\frac{x^4}{4})-5(\frac{x^3}{3})+6(\frac{x^2}{2})+9(x)

=x^4-5\frac{x^3}{3}+3x^2+9x=F(x)

Using second fundamental theorem of calculus

I =F(2)-F(1)

I=\{2^4-\frac{5.(2)^3}{3}+3(2)^2+9(2)\}-\{1^4-\frac{5.(1)^3}{3}+3(1)^2+9(1)\}

=( \,16-\frac{40}{3}+12+18) \,-( \,1-\frac{5}{3}+3+9) \,

=\,16-\frac{40}{3}+12+18 \,- \,1+\frac{5}{3}-3-9 \,

=33-\frac{35}{3}

=\frac{99-35}{3}

=\frac{64}{3}

Question 4: \int_{0}^{\frac{\pi}{4}} sin2x \,dx

Answer:

Let I=\int_{0}^{\frac{\pi}{4}} sin2x \,dx

\int sin2x \,dx = ( \,\frac{-cos2x}{2}) \,=F(x)

Using second fundamental theorem of calculus

I = F(\frac{\pi}{4})-F(0)

=-\frac{1}{2}[ \,cos2( \,\frac{\pi}{4}) \,-cos0] \,

=-\frac{1}{2}[ \,cos( \,\frac{\pi}{2}) \,-cos0] \,

=-\frac{1}{2}[0-1]

=\frac{1}{2}

Question 5: \int_{0}^{\frac{\pi}{2}} cos2x \,dx

Answer:

Let I=\int_{0}^{\frac{\pi}{2}} cos2x \,dx

\int cos2x \,dx = \frac{sin2x}{2} = F(x)

Using second fundamental theorem of calculus

I = F(\frac{\pi}{2})-F(0)

=\frac{1}{2}[ \,sin2( \,\frac{\pi}{2}) \,-sin0] \,

=\frac{1}{2}[ \,sin\pi \,-sin0] \,

=\frac{1}{2}[0-0]=0

Question 6: \int_{4}^{5} e^x\,dx

Answer:

Let I=\int_{4}^{5} e^x\,dx

\int e^x dx = e^x = F(x)

Using second fundamental theorem of calculus

I=F(5)-F(4)

=e^5-e^4

=e^4(e-1)

Question 7: \int_{0}^{\frac{\pi}{4}} tanx \,dx

Answer:

Let I=\int_{0}^{\frac{\pi}{4}} tanx \,dx

\int tanx \,dx=-log|cosx|=F(x)

Using second fundamental theorem of calculus

I=F(\frac{\pi}{4})-F(0)

=-log|cos\frac{\pi}{4}|+log|cos0|

=-log|\frac{1}{\sqrt{2}}|+log|1|

=-log(2)^{-\frac{1}{2}}

=\frac{1}{2}log2

Question 8: \int_{{\frac{\pi}{6}}}^{\frac{\pi}{4}} cosecx \,dx

Answer:

Let I = \int_{{\frac{\pi}{6}}}^{\frac{\pi}{4}} cosecx \,dx

\int cosecx dx= log|cosecx-cotx|=F(x)

Using second fundamental theorem of calculus

I=F(\frac{\pi}{4})-F(\frac{\pi}{6})

=log|cosec\frac{\pi}{4}-cot\frac{\pi}{4}|-log|cosec\frac{\pi}{6}-cot\frac{\pi}{6}|

=log|\sqrt{2}-1|-log|2-\sqrt{3}|

=log( \,\frac{\sqrt{2}-1}{2-\sqrt{3}}) \,

Question 9: \int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}

Answer:

Let I=\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}

\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x = F(x)

Using second fundamental theorem of calculus

I = F(1)-F(0)

= sin^{-1}(1)- sin^{-1}(0)

=\frac{\pi}{2}-0

=\frac{\pi}{2}

Question 10: \int_{0}^{1} \frac{dx}{1+x^2}

Answer:

let\, I = \int_{0}^{1} \frac{dx}{1+x^2}\\ \int \frac{dx}{1+x^2}=tan^{-1}x=F(x)

Using second fundamental theorem of calculus

I = F(1)-F(0)

= tan^{-1}(1)-tan^{-1}(0)

=\frac{\pi}{4}

Question 11: \int_{2}^{3} \frac{dx}{x^2-1}

Answer:

Let I=\int_{2}^{3} \frac{dx}{x^2-1}

\int\frac{dx}{x^2-1}=\frac{1}{2}log|\frac{x-1}{x+1}|=F(x)

Using second fundamental theorem of calculus

I = F(3)-F(2)

=\frac{1}{2}[ \,log|\frac{3-1}{3+1}|-log|\frac{2-1}{2+1}|] \,

=\frac{1}{2}[ \,log|\frac{2}{4}|-log|\frac{1}{3}|] \,

=\frac{1}{2}[ \,log|\frac{1}{2}|-log|\frac{1}{3}|] \,

=\frac{1}{2}[ \,log|\frac{3}{2}|] \,

Question 12: \int_{0}^{\frac{\pi}{2}} cos^2x \,dx

Answer:

Let I = \int_{0}^{\frac{\pi}{2}} cos^2x \,dx

\int cos^2x \,dx =\int( \,\frac{1+cos2x}{2}) \,dx

\frac{x}{2}+\frac{sin2x}{4} = \frac{1}{2}( \,x+\frac{sin2x}{2}) \,=F(x)

Using second fundamental theorem of calculus

I=[ \,F(\frac{\pi}{2})-F(0)] \,

=\frac{1}{2}[ \,( \,\frac{\pi}{2}-\frac {sin\pi}{2}) \,-( \,0+\frac{sin0}{2}) \,] \,

=\frac{1}{2}[ \,\frac{\pi}{2}+0-0-0] \,

=\frac{\pi}{4}

Question 13: \int_{2}^{3} \frac{x\,dx}{x^2+1}

Answer:

Let I=\int_{2}^{3} \frac{x}{x^2+1}dx

\int\frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{2x}{x^2+1}dx

\frac{1}{2}log(1+x^2)=F(x)

Using second fundamental theorem of calculus
I = F(3)-F(2)

=\frac{1}{2}[ \,log(1+(3)^2)-log(1+(2)^2)] \,

=\frac{1}{2}[ \,log(10)-log(5)] \,

=\frac{1}{2}log( \,\frac{10}{5}) \,=\frac{1}{2}log2

Question 14: \int_{0}^{1} \frac{2x+3}{5x^2+1}dx

Answer:

Let I=\int_{0}^{1} \frac{2x+3}{5x^2+1}dx

\int \frac{2x+3}{5x^2+1}dx=\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx

=\frac{1}{5}\int \frac{(10x+15)}{5x^2+1}dx

=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5x^2+1}dx

=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5( \,x^2+\frac{1}{5}) \,}dx

=\frac{1}{5}log(5x^2+1)+\frac{3}{5}.\frac{1}{\frac{1}{\sqrt{5}}}tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}

=\frac{1}{5}log(5x^2+1)+\frac{3}{\sqrt{5}}tan^{-1}(\sqrt5x) = F(x)

Using second fundamental theorem of calculus

I = F(1)-F(0)

=\{\frac{1}{5}log(5+1)+\frac{3}{\sqrt{5}}tan^{-1}(\sqrt5) \}-\{\frac{1}{5}log(1)+\frac{3}{\sqrt{5}}tan^{-1}(0) \}

=\frac{1}{5}log6+\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}

Question 15: \int_{0}^{1} xe^{x^2}dx

Answer:

Let I=\int_{0}^{1} xe^{x^2}dx

Put x² = t=>2x dx=dt

As x->0,t->0 and as x->1,t->1,

I=\frac{1}{2}\int_{0}^{1} e^tdt

I=\frac{1}{2}\int e^tdt=\frac{1}{2}e^t = F(t)

Using second fundamental theorem of calculus

I = F(1)-F(0)

=\frac{1}{2}e-\frac{1}{2}e^0

=\frac{1}{2}(e-1)

Question 16: \int_{1}^{2} \frac{5x^2}{x^2+4x+3}

Answer:

Let I=\int_{1}^{2} \frac{5x^2}{x^2+4x+3}dx

Dividing\,5x^2 \,byx^2+4x+3,we \,obtain

I=\int_{1}^{2} \{5-\frac{20x+15}{x^2+4x+3}\}dx

I=\int_{1}^{2} 5dx-\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx

I={[ \,5x]}_{1}^{2}-\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx

I={5-I}_1,where \,I =\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx \,\,.........(i)

Consider {I}_1=\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx

Let\,20x+15=A\frac{d}{dx}(x^2+4x+3)+B

=2Ax+(4A+B)

Equating the coefficient of x and constant term

A = 10 and B = -25

=> I_1=10\int_{1}^{2} \frac{2x+4}{x^2+4x+3}dx-25\int_{1}^{2} \frac{1}{x^2+4x+3}dx

Let\, x^2+4x+3=t\\=>(2x+4)dx=dt\\=> I_1=10\int\frac{dt}{t}-25\int\frac{dx}{(x^2+2)^2-1^2}

=10logt-25[ \,\frac{1}{2}log( \,\frac{x+2-1}{x+2+1})] \,

=[ \,10log(x^2+4x+3)]_{1}^{2} \,-25[ \,\frac{1}{2}log( \,\frac{x+1}{x+3})]_{1}^{2} \,

=[ \,log15-10log8] \,-25[ \,\frac{1}{2}log\frac{3}{5}-\frac{1}{2}log\frac{2}{4}] \,

=[ \,log(3*5)-10log(4*2)] \,-\frac{25}{2}[ \,log3-log5-log2+log4] \,

[ \,10+\frac{25}{2}] \,log5+[ \,-10-\frac{25}{2}] \,log4+[ \,10-\frac{25}{2}] \,log3+[ \,-10+\frac{25}{2}] \,log2

=\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}

Substituting the value of I_1 in (1),we obtain
I={5-[ \,\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}}] \, \\ =5-\frac{5}{2}[ \,9log\frac{5}{4}-log\frac{3}{2}] \,

Question 17: \int_{0}^{\frac{\pi}{4}} (2sec^2x+x^3+2)dx

Answer:

Let I=\int_{0}^{\frac{\pi}{4}} (2sec^2x+x^3+2)dx

\int(2sec^2x+x^3+2)dx =2tanx+\frac{x^4}{4}+2x=F(x)

Using second fundamental theorem of calculus

I = F({\frac{\pi}{4}})-F(0)

=[ \,\{2tan\frac{\pi}{4}+\frac{1}{4}(\frac{\pi}{4})^4+2(\frac{\pi}{4})\}-( \,2tan0+0+0) \,] \,

=2tan\frac{\pi}{4}+\frac{\pi ^4}{4^5}+\frac{\pi}{2}

=2+\frac{\pi}{2}+\frac{\pi ^4}{1024}

Question 18:\int_{0}^{\pi} ( \,sin^2\frac{x}{2}-cos^2\frac{x}{2}) \, \,dx

Answer:

Let I=\int_{0}^{\pi} ( \,sin^2\frac{x}{2}-cos^2\frac{x}{2}) \, \,dx

=-\int_{0}^{\pi} ( \,cos^2\frac{x}{2}-sin^2\frac{x}{2}) \, \,dx

\int_{0}^{\pi} cosx \, \,dx

\int cosx \,dx = sinx=F(x)

Using second fundamental theorem of calculus

I=F(\pi)-F(0)

=sin\pi-sin0=>0

Question 19: \int_{0}^{2} \frac{6x+3}{x^2+4}dx

Answer:

Let I=\int_{0}^{2} \frac{6x+3}{x^2+4}dx

\int \frac{6x+3}{x^2+4}dx=3\int \frac{2x+1}{x^2+4}dx

=3\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx

=3log(x^2+4)+\frac{3}{2}tan^{-1}\frac{x}{2}=F(x)

Using second fundamental theorem of calculus

I = F(2)-F(0)

=\{3log(2^2+4)+\frac{3}{2}tan^{-1}\frac{2}{2}\}-\{3log(0^2+4)+\frac{3}{2}tan^{-1}\frac{0}{2}\}

=3log(8)+\frac{3}{2}tan^{-1}1-3log(4)-\frac{3}{2}tan^{-1}0

=3log(8)+\frac{3}{2}\frac{\pi}{4}-3log(4)-0

=3log( \,\frac{8}{4}) \,+\frac{3\pi}{4}

=3log2+\frac{3\pi}{4}

Question 20: \int_{0}^{1} ( \,xe^x+sin \frac{\pi x}{4}) \,dx

Answer:

Let I=\int_{0}^{1} ( \,xe^x+sin \frac{\pi x}{4}) \,dx

\int ( \,xe^x+sin \frac{\pi x}{4}) \,dx

=\int ( \,xe^x)dx +\int (sin \frac{\pi x}{4}) \,dx

=x\int e^x dx-\int \{( \,\frac{d}{dx}x) \,\int e^xdx\}dx+\{\frac{-cos\frac{\pi x}{4}}{\frac{\pi}{4}}\}

=xe^x-\int e^xdx-\frac{4}{\pi}cos\frac{\pi x}{4}

=xe^x-e^x-\frac{4}{\pi}cos\frac{\pi x}{4} = F(x)

Using second fundamental theorem of calculus

I = F(1)-F(0)

=( \,1e^1-e^1-\frac{4}{\pi}cos\frac{\pi }{4}) \,-( \,0e^0-e^0-\frac{4}{\pi}cos\frac{\pi *0}{4}) \,

=e-e-\frac{4}{\pi}( \,\frac{1}{\sqrt{2}}) \,+1+\frac{4}{\pi}

=1+\frac{4}{\pi}-\frac{2\sqrt2}{\pi}

Question 21: \int_{1}^{\sqrt{3}}\frac{dx}{1+x^2}\,equals\\\\(A).\frac{\pi}{3}\\\\(B).\frac{2\pi}{3}\\\\(C).\frac{\pi}{6}\\\\(D).\frac{\pi}{12}

Correct Answer is (D).\frac{\pi}{12}

Question 22: \int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2}\,equals

\\\\(A).\frac{\pi}{6}\\\\(B).\frac{\pi}{12}\\\\(C).\frac{\pi}{24}\\\\(D).\frac{\pi}{4}

Correct Answer is (C).\frac{\pi}{24}

Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.9

Chapter 7– Integrals Exercise 7.9

Question 1: \int_{-1}^{1} (x+1) \,dx

Question 2: \int_{2}^{3} \frac{1}{x} \,dx

Question 3: \int_{1}^{2} (4x^3-5x^2+6x+9) \,dx

Question 4: \int_{0}^{\frac{\pi}{4}} sin2x \,dx

Question 5: \int_{0}^{\frac{\pi}{2}} cos2x \,dx

Question 6: \int_{4}^{5} e^x\,dx

Question 7: \int_{0}^{\frac{\pi}{4}} tanx \,dx

Question 8: \int_{{\frac{\pi}{6}}}^{\frac{\pi}{4}} cosecx \,dx

Question 9: \int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}

Question 10: \int_{0}^{1} \frac{dx}{1+x^2}

Question 11: \int_{2}^{3} \frac{dx}{x^2-1}

Question 12: \int_{0}^{\frac{\pi}{2}} cos^2x \,dx

Question 13: \int_{2}^{3} \frac{x\,dx}{x^2+1}

Question 14: \int_{0}^{1} \frac{2x+3}{5x^2+1}dx

Question 15: \int_{0}^{1} xe^{x^2}dx

Question 16: \int_{1}^{2} \frac{5x^2}{x^2+4x+3}

Question 17: \int_{0}^{\frac{\pi}{4}} (2sec^2x+x^3+2)dx

Question 18:\int_{0}^{\pi} ( \,sin^2\frac{x}{2}-cos^2\frac{x}{2}) \, \,dx

Question 19: \int_{0}^{2} \frac{6x+3}{x^2+4}dx

Question 20: \int_{0}^{1} ( \,xe^x+sin \frac{\pi x}{4}) \,dx

Question 21: \int_{1}^{\sqrt{3}}\frac{dx}{1+x^2}\,equals\\\\(A).\frac{\pi}{3}\\\\(B).\frac{2\pi}{3}\\\\(C).\frac{\pi}{6}\\\\(D).\frac{\pi}{12}

Question 22: \int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2}\,equals

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