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NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.10

NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.10

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.10 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.10

Exercise 7.10

Q1. Evaluate the definite integral : 01xx2+1dx using substitution

Answer. 01xx2+1dx  Let x2+1=t2xdx=dt when x=0,t=1 and when x=1,t=201xx2+1dx=1212dtt 

Q2. Evaluate the definite integral : 0πsinϕcos5ϕdϕ using substitution

Answer. Let I=0π2sinϕcos5ϕdϕ=bπsinϕcos4ϕcosϕdϕ Also, let sinϕ=tcosϕdϕ=dt  When ϕ=0,t=0 and when ϕ=π2,t=1I=01t(1t2)2dt=01t12(1+t42t2)dt =01[t12+t922t52]dt=[t3232+t1121122t7272]01 

Q3. Evaluate the definite integral : 01sin1(2x1+x2)dx using substitution

Answer.  Let I=01sin1(2x1+x2)dx Also, let x=tanθdx=sec2θdθ when x=0,θ=0 and when x=1,θ=π4 I=0π4sin1(2tanθ1+tan2θ)sec2θdθ =0π4sin1(sin2θ)sec2θdθ=0π42θsec2θdθ=20π4θsec2θdθ  Taking θ as first function and sec2θ as second function and integrating by parts, we obtain I=2[θθsec2θdθ{(ddxθ)sec2θdθ}dθ]0π4 I=2[θθsec2θdθ{(ddxθ)sec2θdθ}dθ]0x4=2[θtanθtanθdθ]0π4=2[θtanθ+logcosθ]0π4 

Q4. Evaluate the definite integral : 02xx+2( Put x+2=t2) using substitution

Answer. 02xx+2dx Let x+2=t2dx=2tdt when x=0,t=2 and when x=2,t=2 02xx+2dx=22(t22)t22tdt=222(t22)2dt=222(t42t2)dt=2[t552t33]22 

Q5. Evaluate the definite integral : 0π2sinx1+cos2xdx using substitution

Answer. 0π2sinx1+cos2xdx Let cosx=tsinxdx=dt when x=0,t=1 and when x=π2,t=0 

Q6. Evaluate the definite integral : 02dxx+4x2 using substitution

Answer. 02dxx+4x2=02dx(x2x4)=02dx(x2x+14144) =02dx[(x12)2174]=02dx(172)2(x12)2 Let x12=tdx=dt  When x=0,t=12 and when x=2,t=3202dx(172)2(x12)2=1232dt(172)2t2 =[12(172)log172+t172t]1232 =117[log172+3217232log17212log172+12]=117[log17+3173log17117+1] =117log17+3173×17+1171=117log[17+3+41717+3417] =117log[20+41720417]=117log(5+17517) =117log[(5+17)(5+17)2517]=117log[25+17+10178]]