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Question Number 93959 by mashallah last updated on 16/May/20

∫(tan3x+sec3x)dx=

$$\int\left(\mathrm{tan3x}+\mathrm{sec3x}\right)\mathrm{dx}= \\ $$

Answered by john santu last updated on 16/May/20

∫ ((sin 3x)/(cos 3x)) dx = −(1/3)∫ ((d(cos 3x))/(cos 3x))  = −(1/3)ln ∣cos 3x∣   ∫ sec 3x dx = (1/3)∫ sec 3x d(3x)  = (1/3)ln ∣sec 3x + tan 3x ∣   ∴  = −(1/3) ln ∣cos 3x∣ + (1/3)ln∣sec 3x+tan 3x∣ + c  = (1/3)ln ∣((sec 3x+tan 3x)/(cos 3x))∣ + c

$$\int\:\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{cos}\:\mathrm{3x}}\:\mathrm{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{\mathrm{d}\left(\mathrm{cos}\:\mathrm{3x}\right)}{\mathrm{cos}\:\mathrm{3x}} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{3x}\mid\: \\ $$$$\int\:\mathrm{sec}\:\mathrm{3x}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int\:\mathrm{sec}\:\mathrm{3x}\:\mathrm{d}\left(\mathrm{3x}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{3x}\:+\:\mathrm{tan}\:\mathrm{3x}\:\mid\: \\ $$$$\therefore\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{3x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid\mathrm{sec}\:\mathrm{3x}+\mathrm{tan}\:\mathrm{3x}\mid\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{\mathrm{sec}\:\mathrm{3x}+\mathrm{tan}\:\mathrm{3x}}{\mathrm{cos}\:\mathrm{3x}}\mid\:+\:\mathrm{c}\: \\ $$$$ \\ $$

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