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B
−2b2[log|λ|−aλ]+C
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C
−4b2[log|λ|+aλ]+C
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D
−4b2[log|λ|−aλ]+C
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Solution
The correct option is A−2b2[log|λ|+aλ]+C ∫sin2x(a+bcosx)2dx=∫2sinxcosx(a+bcosx)2dxa+bcosx=t⟹cosx=t−ab⟹−bsinxdx=dt=−2b∫(t−ab)1t2dt=−2b2∫(t−at2)dt=−2b2[∫1tdt−∫at2dt]=−2b2[lnt−at−2+1−2+1]+C=−2b2[lnt+at]+C=−2b2[ln(a+bcosx)+aa+bcosx]+C=−2b2[ln|λ|+aλ]+C