Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Step-by-step explanation:

$\sf \int \frac{cos\theta }{7 + sin\theta - 2 {cos}^{2}\theta } d\theta \\ \\ \sf \longrightarrow \int \frac{cos\theta}{7 + sin\theta - 2( 1-{sin}^{2}\theta)} \: d\theta \\ \\ \sf \longrightarrow \int \frac{cos\theta}{ 7 + sin\theta - 2 + 2{sin}^{2}\theta } \: d\theta \\ \\ \sf \longrightarrow \int \frac{cos\theta}{5+sin\theta + 2{sin}^{2}\theta } \\ \\ \sf Substitute \: sin\theta = t \\ \\ \sf Differentiating \: w.r.t. \: \theta \\ \\ \sf \longrightarrow cos\theta = \frac{dt}{d\theta} \\ \\ \sf \longrightarrow dt = cos\theta . d\theta \\ \\ \sf \longrightarrow \int \frac{dt}{5 + t + 2{t}^{2}} \\ \\ \sf \longrightarrow \int \frac{dt}{\frac{5}{2} + \frac{t}{2} + {t}^{2}} \\ \\ \sf \longrightarrow \int \frac{dt}{\frac{5}{2} - \frac{1}{16} + {t}^{2} + \frac{t}{2} + \frac{1}{16}} \\ \\ \sf \longrightarrow \int \frac{dt}{\frac{40-1}{16} + {(t+\frac{1}{4})}^{2}} \\ \\ \sf \longrightarrow \int \frac{dt}{{(\frac{\sqrt{39}}{4})}^{2} + {(t+\frac{1}{4})}^{2}} \\ \\ \sf \frac{1}{\frac{\sqrt{39}}{4}} {tan}^{-1} \frac{t+\frac{1}{4}}{\frac{\sqrt{39}}{4}} \\ \\ \sf \longrightarrow \frac{4}{\sqrt{39}}{tan}^{-1} \frac{4t+1}{\sqrt{39}} + c \\ \\ \sf Resubstituting \: the \: value \: of \: t \\ \\ \sf \longrightarrow \frac{4}{\sqrt{39}}{tan}^{-1} \frac{4sin\theta+1}{\sqrt{39}} + c \\ \\ \sf So\:the\:correct\:choice\:is\:option\:C \\ \\ \sf \star Formula \: used \star \\ \\ \sf \int \frac{dx}{{a}^{2}+{x}^{2}} = \frac{1}{a}{tan}^{-1}\frac{x}{a} + c$

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