Question

plz help meee......​

Answer 1

GIVEN :–

$\\ \implies \displaystyle \sf \int \dfrac{dx}{(2sin \ x + sec \ x)^4} = A(1 + tan \ x)^{-5} + B(1 + tan \ x)^{-6} + C(1 + tan \ x)^{-7} + k$

TO FIND :–

• A + B + C = ?

SOLUTION :–

• Let –

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{(2sin \ x + sec \ x)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \bigg(2sin \ x + \dfrac{1}{ \cos x} \bigg)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + 1)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + { \sin}^{2}x + { \cos}^{2} x )^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{({ \sin}x + { \cos}x )^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{ \cos^{8} (x) \bigg( \dfrac{{ \sin}(x)}{ \cos(x) } + 1 \bigg )^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \cos^{4} (x) ( \tan x+ 1)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{\sec^{4} (x).dx}{ ( \tan x+ 1)^8}$

• We should write this as –

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \sec}^{2}(x). \sec^{2} (x).dx}{ ( \tan x+ 1)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + { \tan}^{2}(x) \} \sec^{2} (x).dx}{ ( \tan x+ 1)^8}$

• Put 1 + tan(x) = t –

• So that –

$\\ \displaystyle \implies \bf { \sec}^{2}(x).dx = dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {t}^{2}\}. dt}{ ( 1+ t)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {(t - 1)}^{2}\}. dt}{ (t)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {(t - 1)}^{2}}{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {t}^{2} + 1 - 2t }{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ {t}^{2} - 2t + 2 }{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{1}{ {t}^{6} }dt - 2 \int \dfrac{1}{ {t}^{7} } dt + 2 \int \dfrac{1}{ {t}^{8} } dt$

$\\ \displaystyle \implies \bf I = - \dfrac{1}{5 {t}^{5} } + (2) \dfrac{1}{6{t}^{6} } - (2 ) \dfrac{1}{7 {t}^{7} } + k$

• Now replace 't' –

$\\ \displaystyle \implies \bf I = - \dfrac{ {(1 + \tan x) }^{ - 5} }{5} + \dfrac{{(1 + \tan x) }^{ -6}}{3} - \dfrac{2{(1 + \tan x) }^{ - 7}}{7} + k$

• Now compare –

$\\ \implies \bf A = - \dfrac{ 1}{5}$

$\\ \implies \bf B = \dfrac{ 1}{3}$

$\\ \implies \bf C = - \dfrac{2}{7}$

• So that –

$\\ \implies \bf A+ B+ C =- \dfrac{ 1}{5} + \dfrac{ 1}{3} - \dfrac{2}{7}$

$\\ \implies \bf A+ B+ C =- \dfrac{16}{105}$

• Hence Option (D) is correct.

Answer 2

Answer:

$GIVEN :–</p><p></p><p>\begin{gathered} \\ \implies \displaystyle \sf \int \dfrac{dx}{(2sin \ x + sec \ x)^4} = A(1 + tan \ x)^{-5} + B(1 + tan \ x)^{-6} + C(1 + tan \ x)^{-7} + k \\ \end{gathered}⟹∫(2sin x+sec x)4dx=A(1+tan x)−5+B(1+tan x)−6+C(1+tan x)−7+k</p><p></p><p>TO FIND :–</p><p></p><p>• A + B + C = ?</p><p></p><p>SOLUTION :–</p><p></p><p>• Let –</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{dx}{(2sin \ x + sec \ x)^4} \\ \end{gathered}⟹I=∫(2sin x+sec x)4dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \bigg(2sin \ x + \dfrac{1}{ \cos x} \bigg)^4} \\ \end{gathered}⟹I=∫(2sin x+cosx1)4dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + 1)^4} \\ \end{gathered}⟹I=∫(2sin xcosx+1)4cos4x.dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + { \sin}^{2}x + { \cos}^{2} x )^4} \\ \end{gathered}⟹I=∫(2sin xcosx+sin2x+cos2x)4cos4x.dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{({ \sin}x + { \cos}x )^8} \\ \end{gathered}⟹I=∫(sinx+cosx)8cos4x.dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{ \cos^{8} (x) \bigg( \dfrac{{ \sin}(x)}{ \cos(x) } + 1 \bigg )^8} \\ \end{gathered}⟹I=∫cos8(x)(cos(x)sin(x)+1)8cos4x.dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \cos^{4} (x) ( \tan x+ 1)^8} \\ \end{gathered}⟹I=∫cos4(x)(tanx+1)8dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{\sec^{4} (x).dx}{ ( \tan x+ 1)^8} \\ \end{gathered}⟹I=∫(tanx+1)8sec4(x).dx</p><p></p><p>• We should write this as –</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ { \sec}^{2}(x). \sec^{2} (x).dx}{ ( \tan x+ 1)^8} \\ \end{gathered}⟹I=∫(tanx+1)8sec2(x).sec2(x).dx</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + { \tan}^{2}(x) \} \sec^{2} (x).dx}{ ( \tan x+ 1)^8} \\ \end{gathered}⟹I=∫(tanx+1)8{1+tan2(x)}sec2(x).dx</p><p></p><p>• Put 1 + tan(x) = t –</p><p></p><p>• So that –</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf { \sec}^{2}(x).dx = dt\\ \end{gathered}⟹sec2(x).dx=dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {t}^{2}\}. dt}{ ( 1+ t)^8} \\ \end{gathered}⟹I=∫(1+t)8{1+t2}.dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {(t - 1)}^{2}\}. dt}{ (t)^8} \\ \end{gathered}⟹I=∫(t)8{1+(t−1)2}.dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {(t - 1)}^{2}}{ (t)^8} .dt\\ \end{gathered}⟹I=∫(t)81+(t−1)2.dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {t}^{2} + 1 - 2t }{ (t)^8} .dt\\ \end{gathered}⟹I=∫(t)81+t2+1−2t.dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{ {t}^{2} - 2t + 2 }{ (t)^8} .dt\\ \end{gathered}⟹I=∫(t)8t2−2t+2.dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = \int \dfrac{1}{ {t}^{6} }dt - 2 \int \dfrac{1}{ {t}^{7} } dt + 2 \int \dfrac{1}{ {t}^{8} } dt\\ \end{gathered}⟹I=∫t61dt−2∫t71dt+2∫t81dt</p><p></p><p>\begin{gathered} \\ \displaystyle \implies \bf I = - \dfrac{1}{5 {t}^{5} } + (2) \dfrac{1}{6{t}^{6} } - (2 ) \dfrac{1}{7 {t}^{7} } + k\\ \end{gathered}⟹I=−5t51+(2)6t61−(2)7t71+k</p><p></p><p>• Now</p><p></p><p>$

Step-by-step explanation:

too hard