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∫ sec23xdx =(A) \(k+\frac{sec^23x}{3}\)(B) \(k+\frac{tan3x}{3}\)(C) tan3x + k(D) k - tan3x |
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Answer» Correct option is: (B) \(k+\frac{tan3x}{3}\) Use Integration by Substitution.Let u=3x, du=3 dx , then \(\rm dx=\dfrac{1}{3} \) Using u and du above, rewrite ∫ sec2(3x) dx. \(\\\displaystyle \implies\rm\int \dfrac{\sec^{2}u}{3} \, du\) \(\\\displaystyle \implies \rm\dfrac{1}{3}\int \sec^{2}u \, du\) \(\\ \implies \rm \dfrac{\tan u}{3}\) Substitute u=3x back into the original integral. \(\\ \implies \rm\dfrac{\tan{3x}}{3}\) \(\\ \bigstar\quad\boxed{\rm\dfrac{\tan{3x}}{3}+k}\quad\bigstar\) Hence,Correct option is (B) |
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