If ∫dx(2sin⁡x+sec⁡x)4=At5+Bt6+Ct7+k , where t=(1+tan⁡x)−1, then A+B+C=

I=∫sec4⁡xdx2tan⁡x+sec2⁡x4=∫sec4⁡x(1+tan⁡x)8dx=∫1+tan2⁡xsec2⁡xdx(1+tan⁡x)8Put 1+tan⁡x=t. Then sec2⁡xdx=dtThus, I=∫1+(t−1)2t8dt=∫1t6−2t7+2t8dt=t−5−5+2t−06+2t−7+kHence A=−15,B=13,C=−27. ∴ A+B+C=−15+13−27=−21+35−30105=−16105

If ∫dx(2sin⁡x+sec⁡x)4=At5+Bt6+Ct7+k , where t=(1+tan⁡x)−1, then A+B+C=

I=∫sec4⁡xdx2tan⁡x+sec2⁡x4=∫sec4⁡x(1+tan⁡x)8dx=∫1+tan2⁡xsec2⁡xdx(1+tan⁡x)8Put 1+tan⁡x=t. Then sec2⁡xdx=dtThus, I=∫1+(t−1)2t8dt=∫1t6−2t7+2t8dt=t−5−5+2t−06+2t−7+kHence A=−15,B=13,C=−27. ∴ A+B+C=−15+13−27=−21+35−30105=−16105

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If $\int \frac{d x}{(2 \sin x + \sec x)^{4}} = A t^{5} + B t^{6} + C t^{7} + k,$ where $t = (1 + \tan x)^{- 1}, then A + B + C =$

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$\frac{- 86}{105}$

$\frac{- 1}{105}$

$\frac{- 16}{105}$

$\frac{- 26}{105}$

answer is C.

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Detailed Solution

$I = \int \frac{\sec^{4} x d x}{{(2 \tan x + \sec^{2} x)}^{4}} = \int \frac{\sec^{4} x}{(1 + \tan x)^{8}} d x = \int \frac{(1 + \tan^{2} x) \sec^{2} x d x}{(1 + \tan x)^{8}}$

Put $1 + \tan x = t . Then \sec^{2} x d x = d t$

Thus, $I = \int \frac{1 + (t - 1)^{2}}{t^{8}} d t = \int (\frac{1}{t^{6}} - \frac{2}{t^{7}} + \frac{2}{t^{8}}) d t = \frac{t^{- 5}}{- 5} + \frac{2 t^{- 0}}{6} + \frac{2 t}{- 7} + k$

Hence $A = \frac{- 1}{5}, B = \frac{1}{3}, C = \frac{- 2}{7} . ∴ A + B + C = \frac{- 1}{5} + \frac{1}{3} - \frac{2}{7} = \frac{- 21 + 35 - 30}{105} = \frac{- 16}{105}$