Get the answer to "Evaluate the integrals ∫18log4θθdθ"

sagnuhh

Answered question

2021-10-27

Evaluate the integrals

\int_{1}^{8} \frac{\log_{4} θ}{θ} d θ

Leonard Stokes

Skilled2021-10-28Added 98 answers

We have to evaluate

\int_{1}^{8} \frac{\log_{4} θ}{θ} d θ

From property of logarithm we have,

\log_{a} b = \frac{\log b}{\log a}

Using the above property, the given integral can be written as

\int_{1}^{8} \frac{\frac{\log θ}{\log 4}}{θ} d θ = \int_{1}^{8} \frac{\log θ}{\log 4} \cdot \frac{1}{θ} d θ

Let substitute

t = \log θ

\Rightarrow \frac{d t}{d θ} = \frac{1}{θ}

\Rightarrow d t = \frac{1}{θ} d θ

Limit will also change accordingly.
When

θ \to 1

then

t = \log 1 \Rightarrow t = 0

When

θ \to 8

then

t = \log 8

Therefore, substituting the above value and then evaluating

\int_{1}^{8} \frac{\log θ}{\log 4} \cdot \frac{1}{θ} d θ = \int_{0}^{\log 8} \frac{t}{\log 4} d t

= \frac{1}{\log 4} \int_{0}^{\log 8} t d t

= \frac{1}{\log 4} \cdot \frac{1}{2} [\log^{2} 8 - 0]

= \frac{1}{2} \frac{\log^{2} (2^{3})}{{\log 2}^{2}}

= \frac{3}{4} \log 2

Hence, required answer is

\frac{3}{4} \log 2

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