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T 14sin x + cos x- dx16 + 9 sin 2x0 |
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Answer» ong>Step-by-step EXPLANATION: I=∫ 0 4 π
9+16sin2x sinx+cosx
dx Let sinx−cosx=t (cosx+sinx)dx=dt Again (sinx−cosx) =t 2
⇒sin 2 x+cos 2 x−2sinxcosx=t 2
⇒1−sin2x=t 2
⇒sin2x=1−t 2
When x=0⇒t=sin0−cos0=−1 When x= 4 π
⇒t=sin 4 π
−cos 4 π
= 2
1
− 2
1
=0 ∴I=∫ −1 0
9+16(1−t 2 ) dt
=∫ −1 0
9+16−16t 2
dt
=∫ −1 0
25−16t 2
dt
= 16 1
∫ −1 0
16 25
−t 2
dt
= 16 1
⎣ ⎢ ⎢ ⎡
2× 4 5
1
log ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
4 5
−t 4 5
+t
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
⎦ ⎥ ⎥ ⎤
−1 0
= 4 1
[ 10 1
log ∣ ∣ ∣ ∣ ∣
5−4t 5+4t
∣ ∣ ∣ ∣ ∣
] −1 0
= 40 1
[log ∣ ∣ ∣ ∣ ∣
5−0 5+0
∣ ∣ ∣ ∣ ∣
−log ∣ ∣ ∣ ∣ ∣
5+4 5−4
∣ ∣ ∣ ∣ ∣
] = 40 1
[log1−log( 9 1
)] = 40 1
[0+log9] = 40 log9
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