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$\int \frac{5^{x}}{\sqrt{5(-2 x)-5^{(2 x)}}} d x=$
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Verified Answer
The correct answer is:
$\frac{\sin ^{-1}\left(5^{2 x}\right)}{\log 25}+c$
Let
$\begin{aligned} I &=\int \frac{5^{x}}{\sqrt{\left(5^{2 x}\right)^{-1}-5^{2 x}}} d x \\ &=\int \frac{5^{x}}{\sqrt{\left(\frac{1}{5^{2 x}}\right)-5^{2 x}}} d x=\int \frac{5^{x} \cdot 5^{x}}{\sqrt{1-\left(5^{2 x}\right)}} d x \end{aligned}$
Put $5^{2 \mathrm{x}}=\mathrm{t} \Rightarrow(2 \log 5) 5^{2 \mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow 5^{\mathrm{x}} \cdot 5^{\mathrm{x}} \mathrm{dx}=\frac{\mathrm{dt}}{\log 25}$
$\begin{aligned} \therefore \mathrm{I} &=\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}} \times \frac{\mathrm{dt}}{(\log 25)}=\frac{1}{\log 25} \int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^{2}}} \mathrm{dt} \\ &=\frac{1}{\log 25} \sin ^{-1} \mathrm{t}+\mathrm{c}=\frac{1}{\log 25} \sin ^{-1}\left(5^{2 \mathrm{x}}\right)+\mathrm{c} \end{aligned}$
$\begin{aligned} I &=\int \frac{5^{x}}{\sqrt{\left(5^{2 x}\right)^{-1}-5^{2 x}}} d x \\ &=\int \frac{5^{x}}{\sqrt{\left(\frac{1}{5^{2 x}}\right)-5^{2 x}}} d x=\int \frac{5^{x} \cdot 5^{x}}{\sqrt{1-\left(5^{2 x}\right)}} d x \end{aligned}$
Put $5^{2 \mathrm{x}}=\mathrm{t} \Rightarrow(2 \log 5) 5^{2 \mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow 5^{\mathrm{x}} \cdot 5^{\mathrm{x}} \mathrm{dx}=\frac{\mathrm{dt}}{\log 25}$
$\begin{aligned} \therefore \mathrm{I} &=\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}} \times \frac{\mathrm{dt}}{(\log 25)}=\frac{1}{\log 25} \int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^{2}}} \mathrm{dt} \\ &=\frac{1}{\log 25} \sin ^{-1} \mathrm{t}+\mathrm{c}=\frac{1}{\log 25} \sin ^{-1}\left(5^{2 \mathrm{x}}\right)+\mathrm{c} \end{aligned}$
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