Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 21 :
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

$\log_{e} 17 - \log_{e} 18$
$\log_{e} 19 - \log_{e} 20$
$\frac{1}{2} (\log_{e} 19 - \log_{e} 17)$
$\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

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(4)

$Let x^{2} = t, 2 x d x = d x$

$f (x) = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$

$f (x) = \frac{1}{2} \log_{e} (\frac{t + 1}{t + 3}) + c = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}) + c$

$f (3) = \frac{1}{2} \log_{e} (\frac{10}{12}) + c = \frac{1}{2} \log_{e} \frac{5}{6} + c = \frac{1}{2} (\log_{e} 5 - \log_{e} 6) \Rightarrow c = 0$

$f (x) = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}); f (4) = \frac{1}{2} \log_{e} (\frac{17}{19}) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

Q 22 :
Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

0%

(64)

$We have, I (x) = \int \sqrt{\frac{x + 7}{x}} d x$

$Put x = t^{2} \Rightarrow d x = 2 t d t$

$So, \int 2 \sqrt{t^{2} + 7} d t = 2 \int \sqrt{t^{2} + {(\sqrt{7})}^{2}} d t$

$= 2 [\frac{t}{2} \sqrt{t^{2} + 7} + \frac{7}{2} \ln | t + \sqrt{t^{2} + 7} |] + C$

$\Rightarrow I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$ $+ C$

$We have, I (9) = 12 + 7 \ln 7 = 12 + 7 [\ln (3 + 4)]$

$\Rightarrow C = 0$

$So, I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$

$I (1) = \sqrt{8} + 7 \ln$ $| 1 + \sqrt{8} |$

$∴$ $I (1) = \sqrt{8} + 7 \ln$ $| 1 + 2 \sqrt{2} |$ $\Rightarrow α = \sqrt{8}$

$∴$ $α^{4} = {[{(8)}^{1 / 2}]}^{4} = 8^{2} \Rightarrow α^{4} = 64$

Q 23 :
Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

0%

(28)

$Given, f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$

$x = \frac{1}{t} \Rightarrow d x = \frac{- 1}{t^{2}} d t$

$\Rightarrow f (x) = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(3 t^{2} + 4)}{t^{2}} \frac{\sqrt{4 t^{2} - 3}}{t}}$

$= - \int \frac{- t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}}$ $Put 4 t^{2} - 3 = z^{2} \Rightarrow t d t = \frac{z}{4} d z$

$=$ $- \frac{1}{4} \int \frac{z d z}{(3 (\frac{z^{2} + 3}{4}) + 4) z} = \int \frac{- d z}{3 z^{2} + 25} = - \frac{1}{3} \int \frac{d z}{z^{2} + {(\frac{5}{\sqrt{3}})}^{2}}$

$= \int - \frac{1}{3} \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} z}{5}) + C = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{4 t^{2} - 3}) + C$

$f (x) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{\frac{4 - 3 x^{2}}{x^{2}}}) + C$

$∵$ $f (0) = 0$ $∴$ $C = \frac{π}{10 \sqrt{3}}$

$Now, f (1) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5}) + \frac{π}{10 \sqrt{3}}$

$\Rightarrow f (1) = \frac{1}{5 \sqrt{3}} \cot^{- 1} (\frac{\sqrt{3}}{5}) = \frac{1}{5 \sqrt{3}} \tan^{- 1} \frac{5}{\sqrt{3}}$

$So, α = 5 : β = \sqrt{3}$ $∴$ $α^{2} + β^{2} = 28$

Q 24 :
If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

0%

(1)

$Let I = \int \sqrt{(\sec 2 x - 1)} d x = \int \sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} d x$

$= \int \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x - 1}} d x$ $= \sqrt{2} \int \frac{\sin x}{\sqrt{2 \cos^{2} x - 1}} d x$

$Substitute \cos x = t \Rightarrow - \sin x d x = d t$

$I = - \sqrt{2} \int \frac{d t}{\sqrt{2 t^{2} - 1}} = - \ln$ $| \sqrt{2} t + \sqrt{2 t^{2} - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{2 \cos^{2} x - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{\cos 2 x} |$

$= - \frac{1}{2} \ln$ $| 2 \cos^{2} x + \cos 2 x + 2 \sqrt{\cos 2 x} \cdot \sqrt{2} \cos x |$ $+ C$

$= \frac{- 1}{2} \ln (2 \cos 2 x + 1 + 2 \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln (\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln$ $| \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \cdot \sqrt{1 + \cos 2 x} |$ $+ C$

$\Rightarrow - \frac{1}{2} \ln$ $| \cos x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} |$ $+ C$

$= α \ln$ $| \cos 2 x + β |$ $+ \sqrt{\cos 2 x (1 + \cos \frac{1}{β})} + c$

On comparing, we get, $α = - \frac{1}{2}, β = \frac{1}{2}$ $∴$ $β - α = \frac{1}{2} + \frac{1}{2} = 1$

Topic Question Set

Q 21 :
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

Q 22 :
Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

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Q 23 :
Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

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Q 24 :
If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

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Topic Question Set

Q 21 : Let f(x)=∫2x(x2+1)(x2+3)dx. If f(3)=12(loge5-loge6), then f(4) is equal to [2023]

Q 22 : Let I(x)=∫x+7xdx and I(9)=12+7loge7. If I(1)=α+7loge(1+22) then α4 is equal to _______ . [2023]

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Q 23 : Let f(x)=∫dx(3+4x2)4-3x2,|x|<23. If f(0)=0 and f(1)=1αβtan-1(αβ), α,β>0, then α2+β2 is equal to _______ . [2023]

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Q 24 : If ∫sec2x-1 dx=αloge|cos2x+β+cos2x(1+cos1βx)|+constant, then β-α is equal to _________ . [2023]

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Q 21 :
$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

Q 22 :
Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

Q 23 :
Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

Q 24 :
If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]