I=∫dx(3+4x^2)√((4 3x^2)) Put x=2√(3)sin t ⇒√(4 3x^2)=2 cos t ⇒ dx=2√(3)cos t dt ⇒∫2√(3)cos t dt(3+163sin^2t)√(4 4sin^2t) ⇒ I= ∫√(3) dt9+16sin^2t ⇒ I= ∫√(3)^2t d ...

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Integration by Substitution

∫dx3+4x2√4-3x...

Question

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

\frac{1}{\sqrt{75}} {tan}^{- 1} (\frac{5 x}{3 \sqrt{4 - 3 x}}) + c

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\frac{1}{\sqrt{20}} {tan}^{- 1} (\frac{5 x}{3 \sqrt{4 - 3 x}}) + c

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\frac{1}{5 \sqrt{3}} {tan}^{- 1} (\frac{5 x}{\sqrt{12 - 9 x^{2}}}) + c

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\frac{1}{5 \sqrt{3}} {tan}^{- 1} (\frac{5 x}{\sqrt{12 - 9 x^{2}}}) + c

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Solution

The correct option is C $\frac{1}{5 \sqrt{3}} {tan}^{- 1} (\frac{5 x}{\sqrt{12 - 9 x^{2}}}) + c$
$I = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{(4 - 3 x^{2})}}$
Put $x = \frac{2}{\sqrt{3}} sin t \Rightarrow \sqrt{4 - 3 x^{2}} = 2 cos t$
$\Rightarrow d x = \frac{2}{\sqrt{3}} cos t d t$
$\Rightarrow \int \frac{\frac{2}{\sqrt{3}} cos t d t}{(3 + \frac{16}{3} {sin}^{2} t) \sqrt{4 - 4 {sin}^{2} t}}$
$\Rightarrow I = \int \frac{\sqrt{3} d t}{9 + 16 {sin}^{2} t}$
$\Rightarrow I = \int \frac{\sqrt{3} {sec}^{2} t d t}{9 {sec}^{2} t + 16 {tan}^{2} t}$
$\Rightarrow I = \int \frac{\sqrt{3} {sec}^{2} t d t}{9 + 25 {tan}^{2} t}$
$tan t = r \Rightarrow {sec}^{2} t d t = d r$
$\Rightarrow I = \int \frac{\sqrt{3} d r}{9 + 25 r^{2}} = \frac{\sqrt{3}}{9} \int \frac{d r}{1 + {(\frac{5 r}{3})}^{2}} = \frac{\sqrt{3}}{9} {tan}^{- 1} (\frac{5 r}{3}) \times \frac{3}{5}$
$x = \frac{2}{\sqrt{3}} sin t \Rightarrow tan t = \frac{\sqrt{3} x}{\sqrt{4 - 3 x^{2}}}$
$I = \frac{1}{5 \sqrt{3}} {tan}^{- 1} (\frac{5 x}{\sqrt{12 - 9 x^{2}}}) + c$

Suggest Corrections

∫dx(3+4x2)√(4−3x2)=

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