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Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 2

Last Updated : 26 May, 2021
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Evaluate the following definite integrals:

Question 23. \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx

I = \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2(1-cos^2x))dx

I = \int_{0}^{\frac{\pi}{2}} [(a^2-b^2)cos^2x+b^2]dx

I = \int_{0}^{\frac{\pi}{2}} [(a^2-b^2)(\frac{1+cos2x}{2})]dx+\int_{0}^{\frac{\pi}{2}}b^2dx

I = \frac{a^2-b^2}{2}\int_{0}^{\frac{\pi}{2}} [(1+cos2x)]dx+\int_{0}^{\frac{\pi}{2}}b^2dx

I = \frac{a^2-b^2}{2}\left[x+\frac{sin2x}{2}\right]_{0}^{\frac{\pi}{2}}+b^2\left[x\right]_{0}^{\frac{\pi}{2}}

I = \frac{a^2-b^2}{2}\left[\frac{\pi}{2}+sin\pi-0-0\right]+b^2\left[\frac{\pi}{2}-0\right]

I = \frac{a^2-b^2}{2}\left[\frac{\pi}{2}\right]+b^2\left[\frac{\pi}{2}\right]

I = (\frac{a^2-b^2}{2}+b^2)\left[\frac{\pi}{2}\right]

I = (\frac{a^2-b^2+2b^2}{2})\left[\frac{\pi}{2}\right]

I = [(a2 + b2)/2][π/2]

I = π(a2 + b2)/4

Therefore, the value of \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx      is π(a2 + b2)/4.

Question 24. \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{1+tan^2\frac{x}{2}+2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{sec^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \frac{1+tan\frac{x}{2}}{sec\frac{x}{2}}dx

I = \int_{0}^{\frac{\pi}{2}} (cos\frac{x}{2}+sin\frac{x}{2})dx

I = \left[2sin\frac{x}{2}-2cos\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2[sinπ/4 – cosπ/4 – 0 + 1]

I = 2[1/√2 – 1/√2 – 0 + 1]

I = 2 (1)

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx       is 2.

Question 25. \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{2cos^2\frac{x}{2}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{2}cos\frac{x}{2}dx

I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}cos\frac{x}{2}dx

I = \sqrt{2}\left[2sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2\sqrt{2}\left[sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2√2[sinπ/4 – sin0]

I = 2√2[1/√2- sin0]

I = 2√2[1/√2] 

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx      is 2.

Question 26. \int_{0}^{\frac{\pi}{2}} xsinxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} xsinxdx

By using integration by parts, we get,

I = x ∫sinxdx – ∫(∫sin x (1)dx)dx

I = -xcosx – ∫(∫sin xdx)dx

I = -xcosx + ∫cosxdx

I = -xcosx + sinx

So we get,

I = \left[-xcosx+sinx\right]^{\frac{\pi}{2}}_0

I = [-π/2cosπ/2 + sinπ/2 + 0 – 0]

I = 0 + 1 + 0 – 0 

I = 1

Therefore, the value of \int_{0}^{\frac{\pi}{2}} xsinxdx      is 1.

Question 27. \int_{0}^{\frac{\pi}{2}} xcosxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} xcosxdx

By using integration by parts, we get,

I = x∫cosxdx – ∫(∫cos x (1)dx)dx

I = xsinx – ∫(∫cosxdx)dx

I = xsinx – ∫sinxdx

I = x sin x + cos x

So we get,

I = \left[xsinx+cosx\right]^{\frac{\pi}{2}}_0

I = [π/2sinπ/2 + cosπ/2 – 0 – cos0]

I = π/2 + 0 – 0 – 1 

I = π/2 – 1  

Therefore, the value of \int_{0}^{\frac{\pi}{2}} xcosxdx      is π/2 – 1.

Question 28. \int_{0}^{\frac{\pi}{2}} x^2cosxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cosxdx

By using integration by parts, we get,

I = x2sinx – ∫(2x∫(cosx)dx)dx

I = x2sinx – ∫(2xsinx)dx

I = x2sinx – 2[-xcosx – ∫(1∫sinxdx)dx]

I = x2sinx – 2[-xcosx + ∫sinxdx]

I = x2sinx – 2[-xcosx + sinx]

I = x2sinx + 2xcosx – 2sinx

So we get,

I = \left[x^2sinx+2xcosx-2sinx\right]^{\frac{\pi}{2}}_0

I = [(π/2)2sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]

I = [π2/4 + 0 – 2 – 0 – 0 + 0]

I = π2/4 – 2 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cosxdx       is π2/4 – 2.

Question 29. \int_{0}^{\frac{\pi}{4}} x^2sinxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}} x^2sinxdx

By using integration by parts, we get,

I = -x2cosx – ∫(2x∫sinxdx)dx

I = -x2cosx + ∫(2xcosx)dx

I = -x2cosx + 2[xsinx – ∫(∫cosxdx)dx]

I = -x2cosx + 2[xsinx – ∫sinxdx]

I = -x2cosx + 2[xsinx + cosx]

I = -x2cosx + 2xsinx + 2cosx

So we get,

I = \left[-x^2cosx+2xsinx+2cosx\right]^{\frac{\pi}{4}}_0

I = -(π/4)2cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2

I = –π2/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2

I = –π2/16√2 + π/2√2 + √2 –  2

Therefore, the value of \int_{0}^{\frac{\pi}{4}} x^2sinxdx      is -π2/16√2 + π/2√2 + √2 –  2.

Question 30. \int_{0}^{\frac{\pi}{2}} x^2cos2xdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cos2xdx

By using integration by parts, we get,

I = 1/2x2sin2x – ∫(2x∫cos2xdx)dx

I = 1/2x2sin2x – ∫(xsin2x)dx

I = 1/2x2sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]

I = 1/2x2sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x2sin2x – [-1/2xcos2x + 1/4sin2xdx]

I = 1/2x2sin2x + 1/2xcos2x – 1/4sin2xdx

So we get,

I = \left[\frac{1}{2}x^2sin2x+\frac{1}{2}xcos2x-\frac{1}{4}sin2xdx\right]^{\frac{\pi}{2}}_0

I = [1/2(π2/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]

I = -π/4

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cos2xdx      is -π/4.

Question 31. \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx

I = \int_{0}^{\frac{\pi}{2}} x^2(\frac{1+cos2x}{2})dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2+x^2cos2x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} x^2dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2cos2x)dx

By using integration by parts, we get,

I = 1/2[x3/3] + x2sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]

I = 1/2[x3/3] + x2sin2x/2 + xcosx/2 – sin2x/4

So we get,

I = \left[\frac{1}{2}[\frac{x^3}{3}]+\frac{x^2sin2x}{2}+\frac{xcosx}{2}-\frac{sin2x}{4}\right]^{\frac{\pi}{2}}_0

I = [1/6[π3/8] + 0 + 0 – π/8] 

I = π3/48 – π/8

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx       is π3/48 – π/8.

Question 32. \int_{1}^{2}logxdx

Solution:

We have,

I = \int_{1}^{2}logxdx

By using integration by parts, we get,

I = xlogx(1)-\int x\frac{1}{x}dx

I = xlogx – ∫1dx

I = xlogx – x

So we get,

I = \left[xlogx-x\right]^2_1

I = 2log2 – 2 – log1 + 1

I = 2 log 2 – 1

Therefore, the value of \int_{1}^{2}logxdx      is 2 log 2 – 1.

Question 33. \int_{1}^{3}\frac{logx}{(x+1)^2}dx

Solution:

We have,

I = \int_{1}^{3}\frac{logx}{(x+1)^2}dx

By using integration by parts, we get,

I = (logx)\frac{(x+1)^{-2+1}}{-2+1}-\int (\frac{1}{x}\int \frac{1}{(x+1)^2}dx)dx

I = -(x+1)^{-1}logx+\int \frac{1}{x(x+1)}dx

I = -\frac{logx}{x+1}+\int (\frac{1}{x}-\frac{1}{x+1})dx

I = -\frac{logx}{x+1}+logx - log(x+1)

So we get,

I = \left[-\frac{logx}{x+1}+logx - log(x+1)\right]^3_1

I = -log3/4 + log3 – log4 + log1/2 – log1 + log2

I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2

I = 3/4log3 – log2

Therefore, the value of \int_{1}^{3}\frac{logx}{(x+1)^2}dx     is 3/4log3 – log2.

Question 34. \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx

Solution:

We have,

I = \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx

I = \int_{1}^{e}(\frac{e^x}{x}+e^xlogx)dx

I = \int_{1}^{e}\frac{e^x}{x}dx+\int_{1}^{e}e^xlogxdx

By using integration by parts, we get,

I = e^xlogx-\int_{1}^{e}e^xlogxdx+\int_{1}^{e}e^xlogxdx

I = exlogx

So we get,

I = \left[e^xlogx\right]^e_1

I = eeloge – e1log1

I = ee (1) – 0

I = ee

Therefore, the value of \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx     is ee.

Question 35. \int_{1}^{e}\frac{logx}{x}dx

Solution:

We have,

I = \int_{1}^{e}\frac{logx}{x}dx

Let log x = t, so we have,

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = e

=> t = log x

=> t = log e

=> t = 1

So, the equation becomes,

I = \int_{1}^{e}\frac{logx}{x}dx

I = \int_{0}^{1}tdt

I = \left[\frac{t^2}{2}\right]^1_0

I = 1/2 – 0/2

I = 1/2

Therefore, the value of \int_{1}^{e}\frac{logx}{x}dx     is 1/2.

Question 36. \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx

Solution:

We have,

I = \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx

I = \int_{e}^{e^2}\frac{1}{logx}dx-\int_{e}^{e^2}\frac{1}{(logx)^2}dx

By using integration by parts, we get,

I = \frac{x}{logx}-\int [(\frac{-1}{(logx)^2})(\frac{1}{x})\int dx]dx -\int \frac{1}{(logx)^2}dx

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

I = \frac{x}{logx}+\int \frac{1}{(logx)^2}dx -\int \frac{1}{(logx)^2}dx

I = x/logx

So we get,

I = \left[\frac{x}{logx}\right]^{e^2}_e

I = \left[\frac{e^2}{loge^2}-\frac{e}{loge}\right]

I = \left[\frac{e^2}{2loge}-e\right]

I = e2/2 – e

Therefore, the value of \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx     is e2/2 – e.

Question 37. \int_{1}^{2}\frac{x+3}{x(x+2)}dx

Solution:

We have,

I = \int_{1}^{2}\frac{x+3}{x(x+2)}dx

I = \int_{1}^{2}\frac{x}{x(x+2)}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx

I = \int_{1}^{2}\frac{1}{x+2}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx

I = \int_{1}^{2}\frac{1}{x+2}dx+\frac{3}{2}\int_{1}^{2}(\frac{1}{x}-\frac{1}{x+2})dx

I = \left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1

I = \left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1

I = \left[\frac{3}{2}logx-\frac{1}{2}log(x+2)\right]^2_1

I = 1/2[3log2 – log4 + log3]

I = 1/2[3log2 – 2log2 + log3]

I = 1/2[log 2 – log 3]

I = 1/2[log6]

I = log6/2

Therefore, the value of \int_{1}^{2}\frac{x+3}{x(x+2)}dx     is log6/2.

Question 38. \int_{0}^{1}\frac{2x+3}{5x^2+1}dx

Solution:

We have,

I = \int_{0}^{1}\frac{2x+3}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{5(2x+3)}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x+15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}(\frac{10x}{5x^2+1}+\frac{15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{1}{5}\int_{0}^{1}\frac{15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+3\int_{0}^{1}\frac{1}{5(x^2+\frac{1}{5})}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{5}(\frac{1}{\frac{1}{\sqrt{5}}})tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}\right]^1_0

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0

I = [1/5log6 + 3/√5tan-1(√5) – 1/5log1 – 3/√5tan-1(0)]

I = [1/5 log6 + 3√5 tan-1(√5) – 0 – 0]

I = 1/5 log6 + 3√5 tan-1(√5)

Therefore, the value of \int_{0}^{1}\frac{2x+3}{5x^2+1}dx      is 1/5 log6 + 3√5 tan-1(√5).

Question 39. \int_{0}^{2}\frac{1}{x+4-x^2}dx

Solution:

We have,

I = \int_{0}^{2}\frac{1}{x+4-x^2}dx

I = \int_{0}^{2}\frac{1}{-(x^2+x-4)}dx

I = \int_{0}^{2}\frac{1}{-(x^2+x+\frac{1}{4}-4-\frac{1}{4})}dx

I = \int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-\frac{17}{4}}dx

I = \int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-(\frac{\sqrt{17}}{2})^2}dx

I = \int_{0}^{2}\frac{1}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}dx

Let x – 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

So, the equation becomes,

I = \int_{\frac{-1}{2}}^{\frac{3}{2}}\frac{1}{(\frac{\sqrt{17}}{2})^2-t^2}dt

I = \left[\frac{1}{2(\frac{\sqrt{17}}{2})}log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]^{\frac{3}{2}}_{\frac{-1}{2}}

I = \frac{1}{\sqrt{17}}\left[log\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-log\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]

I = \frac{1}{\sqrt{17}}\left[log\frac{\sqrt{17}+3}{\sqrt{17}-3}-log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]

I = \frac{1}{\sqrt{17}}\left[log(\frac{\sqrt{17}+3}{\sqrt{17}-3}×\frac{\sqrt{17}+1}{\sqrt{17}-1})\right]

I = \frac{1}{\sqrt{17}}\left[\log\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]

I = \frac{1}{\sqrt{17}}\left[\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]

I = \frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}

I = \frac{1}{\sqrt{17}}\log\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}

I = \frac{1}{\sqrt{17}}\log\frac{25+17+10\sqrt{17}}{8}

I = \frac{1}{\sqrt{17}}\log\frac{42+10\sqrt{17}}{8}

I = \frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}

Therefore, the value of \int_{0}^{2}\frac{1}{x+4-x^2}dx      is \frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}     .

Question 40. \int_{0}^{1}\frac{1}{2x^2+x+1}dx

Solution:

We have,

I = \int_{0}^{1}\frac{1}{2x^2+x+1}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{x^2+\frac{x}{2}+\frac{1}{2}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{1}{2}-\frac{1}{16}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{7}{16}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+(\frac{\sqrt{7}}{4})^2}dx

I = \left[\frac{1}{2}\frac{4}{\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0

I = \left[\frac{4}{2\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0

I = \frac{4}{2\sqrt{7}}\left[\tan^{-1}(\frac{\frac{5}{4}}{\frac{\sqrt{7}}{4}})-\tan^{-1}(\frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]

I = 4/2√7[tan-1(5/√7) – tan-1(1/√7)]

I = 2/√7[tan-1(5/√7) – tan-1(1/√7)]

Therefore, the value of \int_{0}^{1}\frac{1}{2x^2+x+1}dx     is 2/√7[tan-1(5/√7) – tan-1(1/√7)].

Question 41. \int_{0}^{1}\sqrt{x(1-x)}dx

Solution:

We have,

I = \int_{0}^{1}\sqrt{x(1-x)}dx

Let x = sin2 t, so we have,

=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin2 t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1

=> sin2 t = 1

=> sin t = 1

=> t = π/2

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(1-sin^2t)}(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(cos^2t)}(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}(sintcost)(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}(2sin^2tcos^2t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(4sin^2tcos^2t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(sin^22t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1-cos4t}{2})dt

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos4t)dt

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}dt-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(cos4t)dt

I = \frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{4}\left[\frac{sin4t}{4}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{16}\left[sin4t\right]^{\frac{\pi}{2}}_0

I = 1/4[π/2 – 0] – 1/16[sin2π – 0]

I = 1/4[π/2] – 1/16[0 – 0 ]

I = π/8

Therefore, the value of \int_{0}^{1}\sqrt{x(1-x)}dx     is π/8.

Question 42. \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx

Solution:

We have,

I = \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{3+1-(x^2-2x+1)}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{4-(x^2-2x+1)}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{(2)^2-(x-1)^2}}dx

I = \left[sin^{-1}(\frac{x-1}{2})\right]_{0}^{2}

I = [sin-1(1/2) – sin-1(-1/2)]

I = π/6 -(-π/6)

I = π/6 + π/6

I = π/3

Therefore, the value of \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx      is π/3.

Question 43. \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx

Solution:

We have,

I = \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{4-4+4x-x^2}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{4-(x^2-4x+4)}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{2^2-(x-2)^2}}dx

I = \left[sin^{-1}(\frac{x-2}{2})\right]^4_0

I = \left[sin^{-1}(\frac{4-2}{2})-sin^{-1}(\frac{0-2}{2})\right]

I = [sin-1(2/2) – sin-1(-2/2)]

I = sin-11 – sin-1(-1)

I = π/2 – (-π/2) 

I = π/2 + π/2 

I = π

Therefore, the value of \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx      is π.

Question 44. \int_{-1}^{1}\frac{1}{x^2+2x+5}dx

Solution:

We have,

I = \int_{-1}^{1}\frac{1}{x^2+2x+5}dx

I = \int_{-1}^{1}\frac{1}{x^2+2x+1+4}dx

I = \int_{-1}^{1}\frac{1}{(x+1)^2+2^2}dx

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = \int_{0}^{2}\frac{1}{t^2+2^2}dt

I = \left[\frac{1}{2}tan^{-1}\frac{t}{2}\right]^2_0

I = 1/2tan-12/2 – 1/2tan-10/2

I = 1/2tan-11 – 1/2tan-10

I = 1/2(π/4) – 0

I = π/8

Therefore, the value of \int_{-1}^{1}\frac{1}{x^2+2x+5}dx     is π/8.



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