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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.17

  • Last Updated : 19 Apr, 2021

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

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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