Integrals NCERT Solutions
NCERT Exercise 7.6
Integrate the following functions:
Question: 1. `xsinx`
Solution:
Let, `I=intxsinxdx`
`=x int sinxdx-int[d/dx(x)int sinx dx]dx`
`=x(-cosx)-int 1(-cosx)dx`
`=-x cos x+int cos x dx`
`=-cos x + sin x +C`
Question: 2. `xsin3x`
Solution:
Let, `I=int xsin 3x dx`
`=x(-(cos3x)/3)-int1*((-cos3x)/3)dx`
`=-(xcos3x)/3+1/3intcos3xdx`
`=-(xcos3x)/3+1/3*(sin3x)/3+C`
`=-1/3x cos3x+1/3 sin3x +C`
Question: 3. `x^2e^x`
Solution:
Let, `I=int x^2 e^x dx`
`=x^2(e^x)-int 3x(e^x)dx`
`=x^2e^x-2[x(e)^x-1.e^xdx]`
`=x^2e^x-2xe^x+2c^x+C`
`=e^x(x^2-2x+2)+C`
Question: 4. `x logx `
Solution:
Let, `I=int x log dx`
`=log x int x dx - int [d/dx(logx)int x dx] dx`
`=log x(x^2/2)-int(1/x*x^2/2)dx`
`=x^2/2log x - 1/2 int x dx`
`=x^2/2 logx - 1/2 xx x^2/2 + C`
`=x^2/2 log x - 1/4 x^2 + C`
Question: 5. `x log 2x`
Solution:
Let, `I=int x log 2x dx`
`=(log 2x) * x^2/2 - int1/(2x)*2(x^2/2)dx`
`=x^2/2log|2x|-1/2 intx dx`
`=x^2/2 log |2x|-1/2*x^2/2+C`
`=x^2/2 log|2x|-x^2/4+C`
Question: 6. `x^2 log x`
Solution:
Let, `I= intx^2 log x dx`
`=log |x|(x^3/3)-int1/x(x^3/3)dx`
`=x^3/3 log|x| -1/3 intx^2dx`
`=x^3/3 log|x|-1/3 xx x^3/3 +C`
`=x^3/3 log|x| -x^3/9 +C`
Question: 7. `x sin^-1 x`
Solution:
Let, `I=int xsin^-1 x dx`
`=int sin^-1 x *x dx`
`=sin^-1 x. (x^2/2)-int 1/sqrt(1-x^2) * x^2/2 dx`
`=x^2/2 sin^-1 x -1/2 int x^2/sqrt(1-x^2) dx`
`= x^2/2 sin^-1 x - 1/2 I_1`
Where `I_1 = intx^2/sqrt(1-x^2) dx`
Put `x= sin theta`, so that `dx = cos theta d theta`
`:. I_1 = int sin^2 theta/sqrt(1-sin^2theta) .cos theta d theta`
`= int sin^2 theta / cos theta * cos theta d theta`
`=int sin^2 theta d theta`
`=1/2 int (1-cos 2 theta) d theta`
`=1/2 int d theta - 1/2 int (1-cos 2 theta) d theta`
`= 1/2 theta - 1/2 (sin2theta)/2+C`
`=1/2 theta - 1/2 sin theta cos theta + C_1`
`= 1/2 sin^-1 x - 1/2 x sqrt (1-x^2) + C_1`
[since, `sin theta =x => cos theta` `=sqrt(1-sin^2theta)=sqrt(1-x^2)]`
`:. I = x^2/2 sin^-1 x`
`=-1/2[1/2 sin^-1 x -1/2 x-1/2xsqrt(1-x^2)]+C`
`=1/4 sin^-1 x*(2x^2-1)+(xsqrt(1-x^2))/4+C`
Question: 8. `x tan^-1 x`
Solution:
Let, `I =int x tan^-1 x dx`
`=tan^-1 x (x^2/2)- int 1/(1+x^2)*x^2/2 dx`
`=x^2/2 tan^-1 x-1/2 intx^2/(x^2+1)dx`
`=x^2/2 tan^-1 x - 1/2 int (x^2+1-1)/(1+x^2) dx`
`=x^2/2 tan^-1 x - 1/2 int(1-1/(1+x^2))dx`
`=x^2/2 tan^-1 x-1/2(x-tan^-1 x)+C`
`=x^2/2 tan^-1 x-1/2 x+1/2tan^1x + C`
Question: 9. `xcos^-1x`
Solution:
Let `I=int xcos^-1x dx`
`=int cos^-1 x* x dx`
`=cos^-1 x(x^2/2)-int(-1)/sqrt(1-x^2)(x^2/2)dx`
`=x^2/2cos^-1x +1/2 int x^2/sqrt(1-x^2)dx`
`=x^2/2 cos^-1 x + 1/2 I_1`
Where `I_1=int x^2/sqrt(1-x^2)dx`
Put `x= cos theta`, so that `dx=-sin theta d theta`
`:. I_1=int (cos^2 theta (-sin theta))/sqrt(1-cos^2 theta)d theta`
`=-int cos^2 theta d theta`
`=-1/2 int(1+cos2 theta) d theta`
`=-1/2(theta+(sin2 theta)/2)+C_1`
`=-1/2(theta + 1/2 xx 2 sin theta cos theta) +C_1`
`=-1/2(theta+cos theta sqrt(1-cos^2theta)+C_1`
`=-1/2(cos^-1x+x sqrt(1-x^2)+C_1`
`I=cos^-1x/4(2x^2-1)-x/4sqrt(1-x^2)+C`
Question: 10. `(sin^-1x)^2`
Solution:
Let `I=int (sin^-1x)^2 dx`
Put `sin^-1x = theta`
`=>x=sin theta =>dx= cos theta d theta`
`:. I=int theta^2 cos theta d theta`
`=theta^2 (sin theta) - int 2 theta (sin theta) d theta`
`=theta^2 sin theta - 2 theta sin theta d theta`
`=theta^2 sin theta -2[theta*(-cos theta)-` `int 1* (-cos theta) d theta ]+C`
`=theta^2 sin theta + 2 theta cos theta - 2 int cos theta d theta + C`
`= theta^2 sin theta + 2 theta sqrt(1-sin^2 theta) - 2 sin theta + C`
`=x(sin^-1x)^2+2sin^-1x * sqrt(1-x^2)` `- 2x + C`
Question: 11. `(xcos^-1x)/sqrt(1-x^2)`
Solution:
Let, `I =int (xcos^-1x)/sqrt(1-x^2)dx`
Put `cos^-1x = t`, so that `(-1)/(sqrt(1-x^2))dx = dt`
`:. I=-int t cos t dt`
`=-t[t(sint)-int 1*(sint)dt]`
`=-t sin t+ int sin t dt`
`=-t sin t - cos t + C`
`=-t sqrt(1-cos^2t) - cos t + C`
`=-cos^-1x sqrt(1-x^2) - x + C`
`=-[cos^-1x * sqrt(1-x^2)+x]+C`
Question: 12. `x sec^2x`
Solution:
Let, `I=int x sec^2xdx`
`=x(tanx)-int 1* tan x dx`
`=x tan x + log cos x +C`
Question: 13. `tan^-1x`
Solution:
Let, `I=int tan^-1x dx`
`=int tan^-1x*1dx`
`= tan^-1x*x-int (1/(1+x^2))x dx`
`=x tan^-1 x - 1/2 int (2x)/(1+x^2) dx`
`=x tan^-1 x - 1/2 log|x+x^2|+C`
Question: 14. `x(logx)^2`
Solution:
Let `I=int x(logx)^2 dx`
`=int (logx)^2 * x dx`
`=(logx)^2*x^2/2-int[(2logx)*1/x](x^2/2)dx`
`=x^2/2(logx)^2-int(logx)*xdx`
`=x^2/2(logx)^2` `-[(logx)*x^2/2-int1/x*x^2/2dx]`
`=x^2/2(logx)^2-x^2/2logx+1/2intx dx`
`=x^2/2(logx)^2-x^2/2logx+1/2*x^2/2+C`
`=x^2/2(logx)^2-x^2/2logx +1/4x^2+C`
Question: 15. `(x^2+1)logx`
Solution: `int(x^2+1)log x dx`
Taking `log x` as first function and `(x^2+1)` as second function then
`:. int (x^2+1)logx dx`
`= log x int (x^2+1)dx -` `int[d/dx (logx)*int(x^2+1)dx]dx`
`=logx*(x^3/3+x)-int1/x(x^3/3+x)dx`
`=(x^3/3+x)logx - int(x^2/3+1) dx`
`=(x^3/3+x)log - x^3/9 -x +C`
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