Question
Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
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Answer
$\int2\text{x}\sec^3\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)\text{dx}$ $=\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$ Let $\sec\big(\text{x}^2+3\big)=\text{t}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}\text{ dx}=\text{dt}$Now, $\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$
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