Class 12 NCERT Maths – Chapter 7, Exercise 7.7 Solutions

This exercise is based on formulas for integration, before you start the exercise it is advised that you go through the formulas from the following link. Important integration formulas

Q 1: Integrate \sqrt{4 - x^2} dx

Solution:
Let I = \int \sqrt{4 - x^2} \ dx

We use the standard formula:
\int \sqrt{a^2 - x^2} \ dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C

Here a^2 = 4 \implies a = 2,

so:
I = \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1} \frac{x}{2} + C

I = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \frac{x}{2} + C

Final Answer: \boxed{\int \sqrt{4 - x^2} \ dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \frac{x}{2} + C }

Q 2: Integrate \sqrt{1 - 4x^2}

Solution:
Let I = \int \sqrt{1 - 4x^2} \ dx

I = \int \sqrt{1 - (2x)^2} dx

Let u = 2x \implies du = 2 , dx \implies dx = \frac{du}{2}

Then:
I = \frac{1}{2} \int \sqrt{1 - u^2} du

Using: \int \sqrt{a^2 - u^2} , du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1} \frac{u}{a} + C

Here, a = 1

I = \frac{1}{2} \left[ \frac{u}{2} \sqrt{1 - u^2} + \frac{1}{2} \sin^{-1} u \right] + C

Substitute back u = 2x:

I = \frac{x}{2} \sqrt{1 - 4x^2} + \frac{1}{4} \sin^{-1} (2x) + C

Final Answer:
\boxed{ \int \sqrt{1 - 4x^2} \ dx = \frac{x}{2} \sqrt{1 - 4x^2} + \frac{1}{4} \sin^{-1} (2x) + C }

Q 3: Integrate \sqrt{x^2 + 4x + 6}

Solution:
Let I = \int \sqrt{x^2 + 4x + 6} \ dx

Complete the square:
x^2 + 4x + 6 = x^2 + 4x + 4 + 2 = (x + 2)^2 + 2

I = \int \sqrt{(x + 2)^2 + 2} \ dx

Let t = x + 2 \implies dt = dx

I = \int \sqrt{t^2 + 2} \ dt

Using \int \sqrt{x^2 + a^2} \ dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \log \left| x + \sqrt{x^2 + a^2} \right| + C

with a^2 = 2:

I = \frac{t}{2} \sqrt{t^2 + 2} + \frac{2}{2} \log \left| t + \sqrt{t^2 + 2} \right| + C

Substitute back t = x + 2:
I = \frac{(x + 2)}{2} \sqrt{x^2 + 4x + 6} + \log \left| (x + 2) + \sqrt{x^2 + 4x + 6} \right| + C

Final Answer: \boxed{ \int \sqrt{x^2 + 4x + 6} \ dx = \frac{x + 2}{2} \sqrt{x^2 + 4x + 6} + \log \left| x + 2 + \sqrt{x^2 + 4x + 6} \right| + C }

Q 4: Integrate \sqrt{x^2 + 4x + 1}

Solution:
Let, I = \int \sqrt{x^2 + 4x + 1} \ dx

Complete the square:
x^2 + 4x + 1 = x^2 + 4x +1 + 3 -3 = x^2 + 4x +4 - 3 = (x + 2)^2 - 3

Let t = x + 2 \implies dt = dx

I = \int \sqrt{t^2 - 3} \ dt

Using \int \sqrt{x^2 - a^2} , dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log \left| x + \sqrt{x^2 - a^2} \right| + C

with a^2 = 3:

I = \frac{t}{2} \sqrt{t^2 - 3} - \frac{3}{2} \log \left| t + \sqrt{t^2 - 3} \right| + C

Substitute back t = x + 2:

I = \frac{x + 2}{2} \sqrt{x^2 + 4x + 1} - \frac{3}{2} \log \left| x + 2 + \sqrt{x^2 + 4x + 1} \right| + C

Final Answer: \boxed{ \int \sqrt{x^2 + 4x + 1} \ dx= \frac{x + 2}{2} \sqrt{x^2 + 4x + 1} - \frac{3}{2} \log \left| x + 2 + \sqrt{x^2 + 4x + 1} \right| + C }

Q 5: Integrate \sqrt{1 - 4x - x^2}

Solution:
I = \int \sqrt{1 - 4x - x^2} \ dx

Reorder terms: 1 - 4x - x^2 = -(x^2 + 4x - 1)
Complete the square: -(x^2 + 4x - 1) = -(x^2 +4x +4 - 4 -1)= -((x + 2)^2 - 5)
So: 1 - 4x - x^2 = 5 - (x + 2)^2

I = \int \sqrt{5 - (x + 2)^2} \ dx

Let t = x + 2 \implies dt = dx

I = \int \sqrt{5 - t^2} \ dt

Using \int \sqrt{a^2 - x^2} \ dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C

with a^2 = 5:
I = \frac{t}{2} \sqrt{5 - t^2} + \frac{5}{2} \sin^{-1} \frac{t}{\sqrt{5}} + C

Substitute back t = x + 2:
I = \frac{(x + 2)}{2} \sqrt{1 - 4x - x^2} + \frac{5}{2} \sin^{-1} \frac{(x + 2)}{\sqrt{5}} + C

Final Answer: \boxed{ \int \sqrt{1 - 4x - x^2} \ dx = \frac{x + 2}{2} \sqrt{1 - 4x - x^2} + \frac{5}{2} \sin^{-1} \frac{x + 2}{\sqrt{5}} + C }

Q 6: Integrate \sqrt{x^2 + 4x - 5}

Solution:
I = \int \sqrt{x^2 + 4x - 5} \ dx

Complete the square: x^2 + 4x - 5 = x^2 + 4x + 4 - 4 -5 = (x+2)^2 - 9

Let t = x + 2.
Then x^2 + 4x - 5 = t^2 - 9 and dt = dx.

I = \int \sqrt{t^2 - 9} \ dt

Using \int \sqrt{x^2 - a^2} \ dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log \left| x + \sqrt{x^2 - a^2} \right| + C

with a^2 = 9 :

I = \frac{t}{2}\sqrt{t^2 - 9} - \frac{9}{2}\log\Big|t + \sqrt{t^2 - 9}\Big| + C \

I = \frac{(x + 2)}{2}\sqrt{x^2 + 4x - 5} - \frac{9}{2}\log\Big|(x + 2) + \sqrt{x^2 + 4x - 5}\Big| + C

Final Answer: \boxed{\int \sqrt{x^2 + 4x - 5} \ dx = \dfrac{x + 2}{2}\sqrt{x^2 + 4x - 5} - \dfrac{9}{2}\log\Big|x + 2 + \sqrt{x^2 + 4x - 5}\Big| + C}

Q 7: Integrate \sqrt{1 + 3x - x^2} dx

Solution:
Let I = \int \sqrt{1 + 3x - x^2} dx
Complete the square:
1 + 3x - x^2 = - (x^2 - 3x -1) = - (x^2 - 3x + \frac{9}{4} - \frac{9}{4} -1)

1 + 3x - x^2 = -\Big( \Big(x - \frac{3}{2}\Big)^2 - \frac{13}{4} \Big)= \frac{13}{4} - \Big(x - \frac{3}{2}\Big)^2

I = \int \sqrt {\frac{13}{4} - \Big(x - \frac{3}{2}\Big)^2} \ dx

Let t = x - \frac{3}{2} \implies dt = dx

I = \int \sqrt{\frac{13}{4} - t^2} \ dt

Using: \int \sqrt{a^2 - x^2} \ dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \Big(\frac{x}{a}\Big) + C

with a^2 = \frac{13}{4} \Rightarrow a = \frac{\sqrt{13}}{2}:

I = \frac{t}{2} \sqrt{\frac{13}{4} - t^2} + \frac{13}{8} \sin^{-1}\Big(\frac{2t}{\sqrt{13}}\Big) + C

Substitute back t = x - \frac{3}{2}:

I = \frac{x - \frac{3}{2}}{2} \sqrt{\frac{13}{4} - \Big(x - \frac{3}{2}\Big)^2} + \frac{13}{8} \sin^{-1}\Big(\frac{2x - 3}{\sqrt{13}}\Big) + C

I = \frac{2x - 3}{4} \sqrt{1+ 3x - x^2} + \frac{13}{8} \sin^{-1}\Big(\frac{2x - 3}{\sqrt{13}}\Big) + C

Final Answer: \boxed{ \int \sqrt{1 + 3x - x^2} dx = \frac{(2x - 3)}{4} \sqrt{1+ 3x - x^2} + \frac{13}{8} \sin^{-1}\Big(\frac{2x - 3}{\sqrt{13}}\Big) + C }

Q 8: Integrate \sqrt{x^2 + 3x}

Solution:
Let, I = \int \sqrt{x^2 + 3x} dx

Complete the square: x^2 + 3x = \Big(x + \frac{3}{2}\Big)^2 - \frac{9}{4}

I = \int \sqrt{\Bigx + \frac{3}{2} \Big)^2 - \frac{9}{4}} \ dx

Let t = x + \frac{3}{2} \implies dt = dx

\int \sqrt{t^2 - \frac{9}{4}} \ dt

Using \int \sqrt{x^2 - a^2} \ dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log\Big|x + \sqrt{x^2 - a^2}\Big| + C

with a^2 = \frac{9}{4}:

I = \frac{t}{2} \sqrt{t^2 - \frac{9}{4}} - \frac{9}{8} \log\Big|t + \sqrt{t^2 - \frac{9}{4}}\Big| + C

Substitute Back t = x + \frac{3}{2}:

I = \frac{x + \frac{3}{2}}{2} \sqrt{x^2 + 3x} - \frac{9}{8} \log\Big|x + \frac{3}{2} + \sqrt{x^2 + 3x}\Big| + C

I = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log\Big|x + \frac{3}{2} + \sqrt{x^2 + 3x}\Big| + C

Final Answer: \boxed{ \int \sqrt{x^2 + 3x} dx = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log\Big|x + \frac{3}{2} + \sqrt{x^2 + 3x}\Big| + C }

Q 9: Integrate \sqrt{1 + \dfrac{x^2}{9}}

Solution:
Let I = \int \sqrt{1 + \dfrac{x^2}{9}} \ dx

I = \sqrt{1 + \Big ( \dfrac{x}{3} \Big) ^2 }

Let u = \dfrac{x}{3} \implies dx = 3 du

I = 3 \int \sqrt{1 + u^2} du

Using \int \sqrt{x^2 + a^2} dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \log\Big|x + \sqrt{x^2 + a^2}\Big| + C

with a = 1:

I = 3 \Big[ \frac{u}{2} \sqrt{1 + u^2} + \frac{1}{2} \log\Big|u + \sqrt{1 + u^2}\Big| \Big] + C

Substitute u = \dfrac{x}{3}:

I = \dfrac{x}{2} \sqrt{1 + \dfrac{x^2}{9}} + \dfrac{3}{2} \log\Big| \dfrac{x}{3} + \sqrt{1 + \dfrac{x^2}{9}} \Big| + C

I = \dfrac{x}{2} \dfrac{ \sqrt{9 + x^2}}{3} + \dfrac{3}{2} \log\Big| \dfrac{x}{3} + \dfrac {\sqrt{9 + x^2}}{3} \Big| + C

I = \dfrac{x}{6} \sqrt{9 + x^2} + \dfrac{3}{2} \log\Big| \dfrac{x + \sqrt{9 + x^2}}{3} \Big| + C

I = \dfrac{x}{6} \sqrt{9 + x^2} + \dfrac{3}{2} \log\Big| x + \sqrt{9 + x^2} \Big| - \log {3} + C

Final Answer: \boxed{ \int \sqrt{1 + \dfrac{x^2}{9}} \ dx = \dfrac{x}{6} \sqrt{9 + x^2} + \dfrac{3}{2} \log\Big| x + \sqrt{9 + x^2} \Big| + C }

Q 10 (MCQ): Integrate \sqrt{1 + x^2}

(A) \dfrac{x}{2}\sqrt{1+x^2} + \dfrac{1}{2}\log \Big| x + \sqrt{1+x^2} \Big| + C

(B) \dfrac{2}{3}(1+x^2)^{\tfrac{3}{2}} + C

(C) \dfrac{2}{3}x(1+x^2)^{\tfrac{3}{2}} + C

(D) \dfrac{x^2}{2}\sqrt{1+x^2} + \dfrac{1}{2}x^2 \log \Big| x + \sqrt{1+x^2} \Big| + C

Solution:
Using, \int \sqrt{x^2 + a^2} \ dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \log\Big|x + \sqrt{x^2 + a^2}\Big| + C

with a = 1:
\int \sqrt{1 + x^2} \ dx = \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \log\Big|x + \sqrt{1 + x^2}\Big| + C

Correct option: (A)

Q 11: Integrate \sqrt{x^2 - 8x + 7}

Solution:
Let I = \int \sqrt{x^2 - 8x + 7} \ dx

Complete the square: x^2 - 8x + 7 = x^2 - 8x + 16 -16 +7 = (x - 4)^2 - 9

I = \int \sqrt{(x - 4)^2 - 9} \ dx

Let t = x - 4 \implies dt = dx

I = \int \sqrt{t^2 - 9} dt

Using \int \sqrt{x^2 - a^2} \ dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log\Big| x + \sqrt{x^2 - a^2} \Big| + C

with a^2 = 9:
I = \frac{t}{2} \sqrt{t^2 - 9} - \frac{9}{2} \log\Big| t + \sqrt{t^2 - 9} \Big| + C

Substituting back t = x - 4:

I = \frac{x - 4}{2} \sqrt{x^2 - 8x + 7} - \frac{9}{2} \log\Big| x - 4 + \sqrt{x^2 - 8x + 7} \Big| + C

\boxed{ \int \sqrt{x^2 - 8x + 7} \ dx = \dfrac{x - 4}{2} \sqrt{x^2 - 8x + 7} - \dfrac{9}{2} \log\Big| x - 4 + \sqrt{x^2 - 8x + 7} \Big| + C }

Correct option: (D)