RD Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals

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Here we provide RD Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.17
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals

Evaluate the following integrals:

Question 1. ∫dx/√(2x – x2)

Solution:

We have,

Let I = ∫dx/√(2x – x2)

= ∫dx/√(1 – 1 + 2x – x2)

= ∫dx/√{1 – (x2 – 2x + 1)}

= ∫dx/√{12 – (x – 1)2}

Let x – 1 = q …(1)

= ∫dx/√{12 – (q)2}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= sin-1(q) + C    

Now put the value of q from eq(1), we get

= sin-1(x – 1) + C    

Question 2. ∫dx/√(8 + 3x – x2)

Solution:

We have,

Let I = ∫dx/√(8 + 3x – x2)

Here, (8 + 3x – x2) can be written as 8 – (x2 – 3x + 9/4 – 9/4)

= (8 + 9/4) – (x – 3/2)2

= (41/4) – (x – 3/2)2

= ∫\frac{dx}{(\sqrt{\frac{41}{4}})^2-(x-\frac{3}{2})^2}

Let x – 3/2 =  q …(1)

= ∫\frac{dx}{(\sqrt{\frac{41}{4}})^2-(q)^2}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= sin^{-1}(\frac{q}{\sqrt{\frac{41}{4}}}) + C    

Now put the value of q from eq(1), we get

= sin^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}})+C

= sin-1(2x – 3/√41) + C

Question 3. ∫dx/√(5 – 4x – 2x2)

Solution:

We have,

Let I = ∫dx/√(5 – 4x – 2x2)

= ∫dx/√{2(5/2 – 2x – x2)}

= (1/√2)∫dx/√{5/2 – (x2 – 2x + 1 – 1)}

= (1/√2)∫dx/√{(5/2 + 1) – (x2 – 2x + 1)}

= (1/√2)∫dx/√{7/2 – (x – 1)2}

=\frac{1}{\sqrt{2}}∫\frac{dx}{(\sqrt\frac{7}{2})^2-(x+1)^2}

Let x + 1 = q …(1)

= \frac{1}{\sqrt{2}}∫\frac{dx}{(\sqrt\frac{7}{2})^2-(q)^2}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{q}{\sqrt\frac{7}{2}})+C

Now put the value of q from eq(1), we get

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{x+1}{\sqrt\frac{7}{2}})+C

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{(x+1)\sqrt{2}}{\sqrt7})+C

Question 4. ∫dx/√(3x2 + 5x + 7)

Solution:

We have,

Let I = ∫dx/√(3x2 + 5x + 7)

= ∫dx/√{3(x2 + 5x/3 + 7/3)}

= (1/√3)∫dx/√{5/2-(x2-2x+1-1)}

= \frac{1}{\sqrt3}∫\frac{dx}{\sqrt{x^2+\frac{5x}{3}+(\frac{5}{6})^2-(\frac{5}{6})^2+\frac{7}{3}}}

= \frac{1}{\sqrt3}∫\frac{dx}{\sqrt{(x+\frac{5}{6})^2-\frac{25}{36}+\frac{7}{3}}}

= \frac{1}{\sqrt3}∫\frac{dx}{\sqrt{(x+\frac{5}{6})^2+(\frac{\sqrt{56}}{6})^2}}

= \frac{1}{\sqrt3}log|x+\frac{5}{6}+\sqrt{(x+\frac{5}{6})^2-(\frac{59}{36})}|+C

= \frac{1}{\sqrt3}log|x+\frac{5}{6}+\sqrt{x^2+\frac{5x}{3}+\frac{7}{3}}|+C

Question 5. ∫dx/√{(x – α)(β – x)}

Solution:

We have,

Let I = ∫dx/√{(x – α)(β – x)}

= ∫dx/√(-x2 +αx + βx – αβ)

= ∫dx/√{-x2 + x(α + β) – αβ}

= ∫\frac{dx}{\sqrt{-x^2+2x(\frac{α+β}{2})-(\frac{α+β}{2})^2+(\frac{α+β}{2})^2+αβ}}

= ∫\frac{dx}{\sqrt{-[x-(\frac{α+β}{2})^2]+(\frac{α-β}{2})^2}}

= ∫\frac{dx}{\sqrt{(\frac{α+β}{2})^2-[x-(\frac{α+β}{2})^2]}}

Let x – (α + β)/2 = q …(1)

= ∫\frac{dx}{\sqrt{(\frac{α+β}{2})^2-(q)^2}}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= sin^{-1}\frac{q}{\frac{α-β}{2}}+C

Now put the value of q from eq(1), we get

= sin^{-1}\frac{x-\frac{α+β}{2}}{\frac{α-β}{2}}+C

= sin^{-1}(\frac{2x-α-β}{α-β})+C

Question 6. ∫dx/√(7 – 3x – 2x2)

Solution:

We have,

Let I = ∫dx/√(7 – 3x – 2x2)

= ∫dx/√{2(7/2 – 3x/2 – x2)}

= \frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{7}{2}-[x^2+2x(\frac{3}{4})+(\frac{3}{4})^2-(\frac{3}{4})^2}]}

= \frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{7}{2}+\frac{9}{16}-(x+\frac{3}{4})^2}}

= \frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{65}{16}-(x+\frac{3}{4})^2}}

= \frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{(\frac{\sqrt{65}}{4})^2-(x+\frac{3}{4})^2}}

Let x + 3/2 = q …(1)

= \frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{(\frac{\sqrt{65}}{4})^2-(q)^2}}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{q}{\frac{\sqrt{65}}{4}})+C

Now put the value of q from eq(1), we get

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{x+\frac{3}{4}}{\frac{\sqrt{65}}{4}})+C

= \frac{1}{\sqrt{2}}sin^{-1}(\frac{4x+3}{\sqrt{65}})+C

Question 7. ∫dx/√(16 – 6x – x2)

Solution:

We have,

Let I = ∫dx/√(16 – 6x – x2)

= ∫dx/√{16 – (x2 + 2.3x + 9 – 9)}

= ∫dx/√{25 – (x2 + 2.3x + 9)}

= ∫\frac{dx}{(5)^2-(x+3)^2}

Let x + 3/2 = q …(1)

= ∫\frac{dx}{(5)^2-(q)^2}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= sin^{-1}(\frac{q}{5})+C

Now put the value of q from eq(1), we get

= sin^{-1}(\frac{x+3}{5})+C

Question 8. ∫dx/√(7 – 6x – x2)

Solution:

We have,

Let I = ∫dx/√(7 – 6x – x2)

= ∫dx/√{7-(x2 + 2.3x + 9 – 9)}

= ∫dx/√{16 – (x2 + 2.3x + 9)}

= ∫\frac{dx}{(4)^2-(x+3)^2}

Let x + 3/2 = q …(1)

= ∫\frac{dx}{(4)^2-(q)^2}

As we know that, ∫dx/√(a2 – x2) = sin-1(x/a)

So, 

= sin^{-1}(\frac{q}{4})+C

Now put the value of q from eq(1), we get

= sin^{-1}(\frac{x+3}{4})+C

Question 9. ∫dx/√(5x2 – 2x)

Solution:

We have,

Let I = ∫dx/√(5x2 – 2x)

= ∫dx/√{5(x2 – 2x/5)}

= \frac{1}{\sqrt5}∫\frac{dx}{\sqrt{x^2-\frac{2x}{5}+(\frac{1}{5})^2-(\frac{1}{5})^2}}

= \frac{1}{\sqrt5}∫\frac{dx}{\sqrt{(x-\frac{1}{5})^2-(\frac{1}{5})^2}}

= \frac{1}{\sqrt5}log|x-\frac{1}{5}+\sqrt{(x-\frac{1}{5})^2+(\frac{1}{5})^2}|+C

= \frac{1}{\sqrt5}log|x-\frac{1}{5}+\sqrt{x^2-\frac{2x}{5}}|+C

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