Application Q5

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Application Q5

Developed by Dept. of Math., Malda College

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If ∫sinⁿxdx=-((-sinⁿ⁻¹xcos x)/n)+λI_{n-2} then λ=

2 points

((n+1)/n)

((1/(n+1))

((n-1)/n)

n²

Clear selection

If I_{n}=∫₀^{(π/2)}cosⁿxdx and I_{n}=((n-1)/n)λ then λ=

2 points

I_{n-1}

I_{n-2}

I_{n-3}

Clear selection

I_{n}=∫tanⁿxdx and its reduction formula is I_{n}=((tanⁿ⁻¹x)/(n-1))-I_{n-2} . Then I₃=

2 points

(1/2)tan²x

(1/2)tan²x -logsec x

-1

Clear selection

If ∫cosⁿxdx=((cosⁿ⁻¹x f(x))/n)+((n-1)/n)I_{n-2} then f(x) =

2 points

cos x

sec x

- sin x

sin x

Clear selection

Reduction formula of I_{n}=∫₀^{(π/4)}tanⁿxdx is I_{n+1}-I_{n-1}=(1/n) then ∫₀^{(π/4)}tanⁿxdx =

2 points

(π/4)-((76)/(105))

(π/4)-(4/3)

(π/4)-(2/3)

(π/4)-(1/3)

Clear selection

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