Integrals NCERT Solutions

NCERT Exercise 7.7

Integrate the functions in Exercise 1 to 9

Question: 1. `sqrt(4-x^2)`

Solution: `int sqrt(4-x^2) dx`

`=int sqrt((2)^2-x^2) dx`

`=1/2 x sqrt((2)^2-x^2) + (2)^2/2-sin^-1(x/2)+C`

`=1/2 x sqrt(4-x^2)+2sin^-1(x/2)+C`

Question: 2. `sqrt(1-(4x)^2)`

Solution: `int sqrt(1-(4x)^2) dx`

`=2 int sqrt(1/4-x^2) dx`

`=2 int sqrt((1/2)^2 -(x)^2)`

`=2[1/2 x sqrt((1/2)^2 -x^2)+` `(1/2)^2/2 sin^-1 * x/(1//2)]+C`

`=2[1/4 x sqrt(1-4x^2)+1/8 sin^-1(2x)]+C`

`=1/2 x sqrt(1-4x^2)+1/4sin^-1(2x)+C`

Question: 3. `sqrt(x^2+4x+6)`

Solution: `int sqrt(x^2+4x+6)dx`

`=int sqrt(x^2+4x+4+2)dx`

`=int sqrt((x+2)^2+(sqrt 2)^2)dx`

Put `x+2=t =>dx=dt`

`:. int sqrt((x+2)^2+(sqrt 2)^2)dx`

`=int sqrt(t^2 + (sqrt 2)^2)dt`

`=1/2 t sqrt(t^2+(sqrt2)^2) +(sqrt2)^2/2` `log|t+sqrt (t^2+(sqrt2)^2)|+C`

`:.int sqrt(x^2+4x+6)dx`

`=1/2 (x+2) sqrt(x^2+4x+6)+` `log|(x+2)+sqrt(x^2+4x+6)|+C`

Question:4. `sqrt (x^2+4x+1)`

Solution: `int sqrt (x^2+4x+1)dx`

`=int sqrt(x^2+4x+4-3)dx`

`=int sqrt((x+2)^2-(sqrt3)^2)dx`

Put `x+2=t => dx=dt`

`:. int sqrt((x+2)^2-(sqrt3)^2)dx`

`=int sqrt (t^2(sqrt3)^2) dt`

`=t/2 sqrt (t^2(sqrt3)^2) -(sqrt 3)^2/2` `log |t+sqrt(t^2-(sqrt3)^2)|+C`

`:.int sqrt (x^2+4x+1)dx`

`=(x+2)/2 sqrt((x+2)^2-(sqrt3)^2)-3/2log|x+` `2+sqrt((x+2)^2)-(sqrt3)^2)|+C`

`=(x+2)/2 sqrt(x^2+4x+1)` `3/2log|x+2+sqrt(x^2+4x+1)|+C`

Question: 5. `sqrt (1-4x-x^2)`

Solution: `int sqrt (1-4x-x^2)dx`

`=int sqrt(-(x^2+4x-1))dx`

`= int sqrt(-(x^2+4x+4-5)) dx`

`= int sqrt(-[(x+2)^2-(sqrt5)^2])dx`

`= int sqrt((sqrt5)^2-(x+2)^2)dx`

put`x+2=t=>dx=dt`

`=>int sqrt((sqrt5)^2-t^2)dt`

`=1/2t sqrt((sqrt5)^2-t^2)+(sqrt5)^2/2 sin^-1(t/sqrt5)+C`

`:. int sqrt(1-4x-x^2)dx`

`=1/2(x+2)sqrt((sqrt5)^2-(x+2)^2)+` `5/2sin^-1((x+2)/sqrt5)+C`

`=1/2(x+2)sqrt(1-4x-x^2)+` `5/2sin^-1((x+2)/sqrt5)+C`

Question: 6. `sqrt(x^2) + dx - 5)`

Solution: `int sqrt(x^2) + dx - 5)dx`

`=int sqrt(x^2+4x+4-9)dx`

`int sqrt((x+2)^2-(3)^2)dx`

Put `x+2=t => dx =dt`

`:. int sqrt((x+2)^2-(3)^2)dx`

`= int sqrt(t^2-(3)^2)dx`

=`t/2 sqrt(t^2-(3)^2) -` `(3)^2/2 - log|t+sqrt(t^2-(3)^2)|+C`

`:. int sqrt(x^2+4x-5)dx`

`=(x+2)/2 sqrt((x+2)^2-(3)^2)-` `9/2 log|x+2+sqrt((x-2)^2-(3)^2)|+C`

`=(x+2)/2 int sqrt(x^2+4x-5)-` `9/2log|x+2+sqrt(x^2+4x-5)|+C`

Question: 7. `sqrt(1+3x-x^2)`

Solution: `int sqrt(1+3x-x^2)dx`

`=int sqrt(-(x^2-3x-1))dx`

`= int sqrt(-(x^2-3x+9/4-13/4))dx`

`= int sqrt(-[(x-3/2)^2-(sqrt13/2)^2])dx`

`= int sqrt((sqrt 13/2)^2-(x-3/2)^2)dx`

Put `x-3/2 =t=>dx=dt`

`:. int sqrt((sqrt 13/2)^2-(x-3/2)^2)dx`

`=int sqrt ((sqrt 13/2)^2-t^2)dt`

`= 1/2 t sqrt((sqrt13/2)^2-t^2)+` `(sqrt13/2)^2/2 sin^-1* t/(sqrt13//2)+C`

`:. int sqrt(1+3x-x^2)dx`

`= 1/2(x-3/2)sqrt((sqrt13/2)^2-(x-3/2)^2)+` `13/8 sin^-1*((x-3/2))/(sqrt13//2)+C`

`= 1/4(2x-3)sqrt(1+3x-x^2)+` `13/8 sin^-1*((2x-3))/sqrt13 +C`

Question: 8. `sqrt(x^2+3x)`

Solution: `int sqrt(x^2+3x)dx`

`=int sqrt(x^2+3x+9/4-9/4)dx`

`= int sqrt ((x+3/2)^2)-(3/2)^2)dx`

Put `x+3/2=t =>dx=dt`

`:. int sqrt((x+3/2)^2-(3/2)^2)dx`

`=int sqrt(t^2-(3/2)^2)dx`

`=t/2 sqrt(t^2-(3/2)^2)+` `(3/2)^2/2|t+sqrt(t^2-(3/2)^2)|+C`

`:. int sqrt(x^2+3x)dx`

`=1/2(x+3/2)sqrt ((x+3/2)^2-(3/2)^2)-` `9/2log|(x+3/2)+sqrt((x+3/2)^2-(3/2)^2)|+C`

`=1/4(2x+3)sqrt(x^2+3x)-` `9/8log|x+3/2+sqrt(x^2+3x)|+C`

Question: 9. `sqrt(1+x^2/9)`

Solution: `int sqrt(1+x^2/9)dx`

`= 1/3 int sqrt(9+x^2)dx`

`=1/3 int sqrt(3^2+x^2)dx`

`=1/3[1/2 x sqrt(3^2+x^2)+` `3^2/2 log|x+sqrt(3^2+x^2)|]+C`

`=1/6 x sqrt(9+x^2)+` `1/3 xx 9/2 log|x+sqrt(9+x^2)|+C`

` =1/6 x sqrt (9+x^2)+` `3/2 log|x+sqrt(9+x^2)|+C`

Choose the correct answer in Exercises 10 and 11

Question: 10. `int sqrt(1+x^2)dx` is equals to

(A) `x/2 sqrt(1+x^2)+` `1/2 log|x+sqrt(1+x^2)|+C`

(B) `2/3 (1+x^2)^(3/2)+C`

(D) `x^2/2 sqrt(1+x^2)+` `1/2x^2log|x+sqrt(1+x^2)|+C`

Answer: Option (B) `2/3 (1+x^2)^(3/2)+C`

Explanation: `int sqrt(1+x)dx`

`= int sqrt(1^2+x^2)dx`

`=1/2xsqrt(1^2+x^2)+` `1^2/2 log|x+sqrt(1^2+x^2)|+C`

`=1/2x sqrt(1+x^2)+1/2 log|x+sqrt(1+x^2)|+C`

Thus, Option (B) `2/3 (1+x^2)^(3/2)+C` is correct answer

Question: 11. `int sqrt(x^2-8x+7) dx` is equals to

(A) `1/2(x-4)sqrt(x^2-8x+7)+` `9log|x-4+sqrt(x^2-8x+7)|+C`

(B) `1/2(x+4)sqrt(x^2-8x+7)+` `9log|x+4+sqrt(x^2-8x+7)|+C`

(D) `1/2(x-4)sqrt(x^2-8x+7)+` `9/2 log|x-4+sqrt(x^2-8x+7)|+C`

Answer: Option (D)`1/2(x-4)sqrt(x^2-8x+7)+` `9/2 log|x-4+sqrt(x^2-8x+7)|+C`

Explanation:

`int sqrt(x^2-8x+7) dx`

`= int sqrt(x^2-8x+16-9)dx`

`= int sqrt((x-4)^2-3^2)dx`

Put, `x-4=t => dx=dt`

`:. int sqrt((x-4)^2-3^2)dx`

`= int sqrt(t^2- 3^2)dx`

`=t/2 sqrt(t^2-3^2)-` `3^2/2 log|t+sqrt(t^2-3^2)|+C`

`:. int sqrt(x^2-8x+7) dx`

`= 1/2 (x-4) sqrt((x-4)^2-3^2)-` `9/2log|x-4+sqrt((x-4)^2-3^2)|+C`

`=1/2(x-4)sqrt(x^2-8x+7)-` `9/2 log|x-4+sqrt(x^2-8x+7)|+C`

Thus, Option (D) is the correct option

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